Develop Reference

Hackerrank: Climbing the Leaderboard

i have a deal with a hackerrank algorithm problem.
It works at all cases, except 6-7-8-9. It gives timeout error. I had spent so much time at this level. Someone saw where is problem?
static long[] climbingLeaderboard(long[] scores, long[] alice)
{
//long[] ranks = new long[scores.Length];
long[] aliceRanks = new long[alice.Length]; // same length with alice length
long lastPoint = 0;
long lastRank;
for (long i = 0; i < alice.Length; i++)
{
lastPoint = scores[0];
lastRank = 1;
bool isIn = false; // if never drop in if statement
for (long j = 0; j < scores.Length; j++)
{
if (lastPoint != scores[j]) //if score is not same, raise the variable
{
lastPoint = scores[j];
lastRank++;
}
if (alice[i] >= scores[j])
{
aliceRanks[i] = lastRank;
isIn = true;
break;
}
aliceRanks[i] = !isIn & j + 1 == scores.Length ? ++lastRank : aliceRanks[i]; //drop in here
}
}
return aliceRanks;
}

This problem can be solved in O(n) time, no binary search needed at all. First, we need to extract the most useful piece of data given in the problem statement, which is,
The existing leaderboard, scores, is in descending order.
Alice's scores, alice, are in ascending order.
An approach that makes this useful is to create two pointers, one at the start of alice array, let's call it "i", and the second is at the end of scores array, let's call it "j". We then loop until i reaches the end of alice array and at each iteration, we check for three main conditions. We increment i by one if alice[i] is less than scores[j] because the next element of alice may be also less than the current element of scores, or we decrement j if alice[i] is greater than scores[j] because we are sure that the next elements of alice are also greater than those elements discarded in scores. The last condition is that if alice[i] == scores[j], we only increment i.
I solved this question in C++, my goal here is to make you understand the algorithm, I think you can easily convert it to C# if you understand it. If there are any confusions, please tell me. Here is the code:
// Complete the climbingLeaderboard function below.
vector<int> climbingLeaderboard(vector<int> scores, vector<int> alice) {
int j = 1, i = 1;
// this is to remove duplicates from the scores vector
for(i =1; i < scores.size(); i++){
if(scores[i] != scores[i-1]){
scores[j++] = scores[i];
}
}
int size = scores.size();
for(i = 0; i < size-j; i++){
scores.pop_back();
}
vector<int> ranks;
i = 0;
j = scores.size()-1;
while(i < alice.size()){
if(j < 0){
ranks.push_back(1);
i++;
continue;
}
if(alice[i] < scores[j]){
ranks.push_back(j+2);
i++;
} else if(alice[i] > scores[j]){
j--;
} else {
ranks.push_back(j+1);
i++;
}
}
return ranks;
}
I think this may help you too:
vector is like an array list that resizes itself.
push_back() is inserting at the end of the vector.
pop_back() is removing from the end of the vector.

Here is my solution with c#
public static List<int> climbingLeaderboard(List<int> ranked, List<int> player)
{
List<int> result = new List<int>();
ranked = ranked.Distinct().ToList();
var pLength = player.Count;
var rLength = ranked.Count-1;
int j = rLength;
for (int i = 0; i < pLength; i++)
{
for (; j >= 0; j--)
{
if (player[i] == ranked[j])
{
result.Add(j + 1);
break;
}
else if(player[i] < ranked[j])
{
result.Add(j + 2);
break;
}
else if(player[i] > ranked[j]&&j==0)
{
result.Add(1);
break;
}enter code here
}
}
return result;
}

Here is a solution that utilizes BinarySearch. This method returns the index of the searched number in the array, or if the number is not found then it returns a negative number that is the bitwise complement of the index of the next element in the array. Binary search only works in sorted arrays.
public static int[] GetRanks(long[] scores, long[] person)
{
var defaultComparer = Comparer<long>.Default;
var reverseComparer = Comparer<long>.Create((x, y) => -defaultComparer.Compare(x, y));
var distinctOrderedScores = scores.Distinct().OrderBy(i => i, reverseComparer).ToArray();
return person
.Select(i => Array.BinarySearch(distinctOrderedScores, i, reverseComparer))
.Select(pos => (pos >= 0 ? pos : ~pos) + 1)
.ToArray();
}
Usage example:
var scores = new long[] { 100, 100, 50, 40, 40, 20, 10 };
var alice = new long[] { 5, 25, 50, 120 };
var ranks = GetRanks(scores, alice);
Console.WriteLine($"Ranks: {String.Join(", ", ranks)}");
Output:
Ranks: 6, 4, 2, 1

I was bored so i gave this a go with Linq and heavily commented it for you,
Given
public static IEnumerable<int> GetRanks(long[] scores, long[] person)
// Convert scores to a tuple
=> scores.Select(s => (scores: s, isPerson: false))
// convert persons score to a tuple and concat
.Concat(person.Select(s => (scores: s, isPerson: true)))
// Group by scores
.GroupBy(x => x.scores)
// order by score
.OrderBy(x => x.Key)
// select into an indexable tuple so we know everyones rank
.Select((groups, i) => (rank: i, groups))
// Filter the person
.Where(x => x.groups.Any(y => y.isPerson))
// select the rank
.Select(x => x.rank);
Usage
static void Main(string[] args)
{
var scores = new long[]{1, 34, 565, 43, 44, 56, 67};
var alice = new long[]{578, 40, 50, 67, 6};
var ranks = GetRanks(scores, alice);
foreach (var rank in ranks)
Console.WriteLine(rank);
}
Output
1
3
6
8
10

Based on the given constraint brute-force solution will not be efficient for the problem.
you have to optimize your code and the key part here is to look up for exact place which can be effectively done by using binary search.
Here is the solution using binary search:-
static int[] climbingLeaderboard(int[] scores, int[] alice) {
int n = scores.length;
int m = alice.length;
int res[] = new int[m];
int[] rank = new int[n];
rank[0] = 1;
for (int i = 1; i < n; i++) {
if (scores[i] == scores[i - 1]) {
rank[i] = rank[i - 1];
} else {
rank[i] = rank[i - 1] + 1;
}
}
for (int i = 0; i < m; i++) {
int aliceScore = alice[i];
if (aliceScore > scores[0]) {
res[i] = 1;
} else if (aliceScore < scores[n - 1]) {
res[i] = rank[n - 1] + 1;
} else {
int index = binarySearch(scores, aliceScore);
res[i] = rank[index];
}
}
return res;
}
private static int binarySearch(int[] a, int key) {
int lo = 0;
int hi = a.length - 1;
while (lo <= hi) {
int mid = lo + (hi - lo) / 2;
if (a[mid] == key) {
return mid;
} else if (a[mid] < key && key < a[mid - 1]) {
return mid;
} else if (a[mid] > key && key >= a[mid + 1]) {
return mid + 1;
} else if (a[mid] < key) {
hi = mid - 1;
} else if (a[mid] > key) {
lo = mid + 1;
}
}
return -1;
}
You can refer to this link for a more detailed video explanation.

static int[] climbingLeaderboard(int[] scores, int[] alice) {
int[] uniqueScores = IntStream.of(scores).distinct().toArray();
int [] rank = new int [alice.length];
int startIndex=0;
for(int j=alice.length-1; j>=0;j--) {
for(int i=startIndex; i<=uniqueScores.length-1;i++) {
if (alice[j]<uniqueScores[uniqueScores.length-1]){
rank[j]=uniqueScores.length+1;
break;
}
else if(alice[j]>=uniqueScores[i]) {
rank[j]=i+1;
startIndex=i;
break;
}
else{continue;}
}
}
return rank;
}

My solution in javascript for climbing the Leaderboard Hackerrank problem. The time complexity of the problem can be O(i+j), i is the length of scores and j is the length of alice. The space complexity is O(1).
// Complete the climbingLeaderboard function below.
function climbingLeaderboard(scores, alice) {
const ans = [];
let count = 0;
// the alice array is arranged in ascending order
let j = alice.length - 1;
for (let i = 0 ; i < scores.length ; i++) {
const score = scores[i];
for (; j >= 0 ; j--) {
if (alice[j] >= score) {
// if higher than score
ans.unshift(count+1);
} else if (i === scores.length - 1) {
// if smallest
ans.unshift(count+2);
} else {
break;
}
}
// actual rank of the score in leaderboard
if (score !== scores[i-1]) {
count++;
}
}
return ans;
}

Here is my solution
List<int> distinct = null;
List<int> rank = new List<int>();
foreach (int item in player)
{
ranked.Add(item);
ranked.Sort();
ranked.Reverse();
distinct = ranked.Distinct().ToList();
for (int i = 0; i < distinct.Count; i++)
{
if (item == distinct[i])
{
rank.Add(i + 1);
break;
}
}
}
return rank;
This can be modified by removing the inner for loop also
List<int> distinct = null;
List<int> rank = new List<int>();
foreach (int item in player)
{
ranked.Add(item);
ranked.Sort();
ranked.Reverse();
distinct = ranked.Distinct().ToList();
var index = ranked.FindIndex(x => x == item);
rank.Add(index + 1);
}
return rank;

This is my solution in c# for Hackerrank Climbing the Leaderboard based on C++ answer here.
public static List<int> climbingLeaderboard(List<int> ranked, List<int> player)
{
List<int> _ranked = new List<int>();
_ranked.Add(ranked[0]);
for(int a=1; a < ranked.Count(); a++)
if(_ranked[_ranked.Count()-1] != ranked[a])
_ranked.Add(ranked[a]);
int j = _ranked.Count()-1;
int i = 0;
while(i < player.Count())
{
if(j < 0)
{
player[i] = 1;
i++;
continue;
}
if(player[i] < _ranked[j])
{
player[i] = j+2;
i++;
}
else
if(player[i] == _ranked[j])
{
player[i] = j+1;
i++;
}
else
{
j--;
}
}
return player;
}

My solution in Java for climbing the Leaderboard Hackerrank problem.
// Complete the climbingLeaderboard function below.
static int[] climbingLeaderboard(int[] scores, int[] alice) {
Arrays.sort(scores);
HashSet<Integer> set = new HashSet<Integer>();
int[] ar = new int[alice.length];
int sc = 0;
for(int i=0; i<alice.length; i++){
sc = 1;
set.clear();
for(int j=0; j<scores.length; j++){
if(alice[i] < scores[j] && !set.contains(scores[j])){
sc++;
set.add(scores[j]);
}
}
ar[i] = sc;
}return ar;
}

How to random order a 3 digit number on a list box? [duplicate]

What is an elegant way to find all the permutations of a string. E.g. permutation for ba, would be ba and ab, but what about longer string such as abcdefgh? Is there any Java implementation example?

public static void permutation(String str) {
permutation("", str);
}
private static void permutation(String prefix, String str) {
int n = str.length();
if (n == 0) System.out.println(prefix);
else {
for (int i = 0; i < n; i++)
permutation(prefix + str.charAt(i), str.substring(0, i) + str.substring(i+1, n));
}
}
(via Introduction to Programming in Java)

Use recursion.
Try each of the letters in turn as the first letter and then find all the permutations of the remaining letters using a recursive call.
The base case is when the input is an empty string the only permutation is the empty string.

Here is my solution that is based on the idea of the book "Cracking the Coding Interview" (P54):
/**
* List permutations of a string.
*
* #param s the input string
* #return the list of permutations
*/
public static ArrayList<String> permutation(String s) {
// The result
ArrayList<String> res = new ArrayList<String>();
// If input string's length is 1, return {s}
if (s.length() == 1) {
res.add(s);
} else if (s.length() > 1) {
int lastIndex = s.length() - 1;
// Find out the last character
String last = s.substring(lastIndex);
// Rest of the string
String rest = s.substring(0, lastIndex);
// Perform permutation on the rest string and
// merge with the last character
res = merge(permutation(rest), last);
}
return res;
}
/**
* #param list a result of permutation, e.g. {"ab", "ba"}
* #param c the last character
* #return a merged new list, e.g. {"cab", "acb" ... }
*/
public static ArrayList<String> merge(ArrayList<String> list, String c) {
ArrayList<String> res = new ArrayList<>();
// Loop through all the string in the list
for (String s : list) {
// For each string, insert the last character to all possible positions
// and add them to the new list
for (int i = 0; i <= s.length(); ++i) {
String ps = new StringBuffer(s).insert(i, c).toString();
res.add(ps);
}
}
return res;
}
Running output of string "abcd":
Step 1: Merge [a] and b:
[ba, ab]
Step 2: Merge [ba, ab] and c:
[cba, bca, bac, cab, acb, abc]
Step 3: Merge [cba, bca, bac, cab, acb, abc] and d:
[dcba, cdba, cbda, cbad, dbca, bdca, bcda, bcad, dbac, bdac, badc, bacd, dcab, cdab, cadb, cabd, dacb, adcb, acdb, acbd, dabc, adbc, abdc, abcd]

Of all the solutions given here and in other forums, I liked Mark Byers the most. That description actually made me think and code it myself.
Too bad I cannot voteup his solution as I am newbie.
Anyways here is my implementation of his description
public class PermTest {
public static void main(String[] args) throws Exception {
String str = "abcdef";
StringBuffer strBuf = new StringBuffer(str);
doPerm(strBuf,0);
}
private static void doPerm(StringBuffer str, int index){
if(index == str.length())
System.out.println(str);
else { //recursively solve this by placing all other chars at current first pos
doPerm(str, index+1);
for (int i = index+1; i < str.length(); i++) {//start swapping all other chars with current first char
swap(str,index, i);
doPerm(str, index+1);
swap(str,i, index);//restore back my string buffer
}
}
}
private static void swap(StringBuffer str, int pos1, int pos2){
char t1 = str.charAt(pos1);
str.setCharAt(pos1, str.charAt(pos2));
str.setCharAt(pos2, t1);
}
}
I prefer this solution ahead of the first one in this thread because this solution uses StringBuffer. I wouldn't say my solution doesn't create any temporary string (it actually does in system.out.println where the toString() of StringBuffer is called). But I just feel this is better than the first solution where too many string literals are created. May be some performance guy out there can evalute this in terms of 'memory' (for 'time' it already lags due to that extra 'swap')

A very basic solution in Java is to use recursion + Set ( to avoid repetitions ) if you want to store and return the solution strings :
public static Set<String> generatePerm(String input)
{
Set<String> set = new HashSet<String>();
if (input == "")
return set;
Character a = input.charAt(0);
if (input.length() > 1)
{
input = input.substring(1);
Set<String> permSet = generatePerm(input);
for (String x : permSet)
{
for (int i = 0; i <= x.length(); i++)
{
set.add(x.substring(0, i) + a + x.substring(i));
}
}
}
else
{
set.add(a + "");
}
return set;
}

All the previous contributors have done a great job explaining and providing the code. I thought I should share this approach too because it might help someone too. The solution is based on (heaps' algorithm )
Couple of things:
Notice the last item which is depicted in the excel is just for helping you better visualize the logic. So, the actual values in the last column would be 2,1,0 (if we were to run the code because we are dealing with arrays and arrays start with 0).
The swapping algorithm happens based on even or odd values of current position. It's very self explanatory if you look at where the swap method is getting called.You can see what's going on.
Here is what happens:
public static void main(String[] args) {
String ourword = "abc";
String[] ourArray = ourword.split("");
permute(ourArray, ourArray.length);
}
private static void swap(String[] ourarray, int right, int left) {
String temp = ourarray[right];
ourarray[right] = ourarray[left];
ourarray[left] = temp;
}
public static void permute(String[] ourArray, int currentPosition) {
if (currentPosition == 1) {
System.out.println(Arrays.toString(ourArray));
} else {
for (int i = 0; i < currentPosition; i++) {
// subtract one from the last position (here is where you are
// selecting the the next last item
permute(ourArray, currentPosition - 1);
// if it's odd position
if (currentPosition % 2 == 1) {
swap(ourArray, 0, currentPosition - 1);
} else {
swap(ourArray, i, currentPosition - 1);
}
}
}
}

Let's use input abc as an example.
Start off with just the last element (c) in a set (["c"]), then add the second last element (b) to its front, end and every possible positions in the middle, making it ["bc", "cb"] and then in the same manner it will add the next element from the back (a) to each string in the set making it:
"a" + "bc" = ["abc", "bac", "bca"] and "a" + "cb" = ["acb" ,"cab", "cba"]
Thus entire permutation:
["abc", "bac", "bca","acb" ,"cab", "cba"]
Code:
public class Test
{
static Set<String> permutations;
static Set<String> result = new HashSet<String>();
public static Set<String> permutation(String string) {
permutations = new HashSet<String>();
int n = string.length();
for (int i = n - 1; i >= 0; i--)
{
shuffle(string.charAt(i));
}
return permutations;
}
private static void shuffle(char c) {
if (permutations.size() == 0) {
permutations.add(String.valueOf(c));
} else {
Iterator<String> it = permutations.iterator();
for (int i = 0; i < permutations.size(); i++) {
String temp1;
for (; it.hasNext();) {
temp1 = it.next();
for (int k = 0; k < temp1.length() + 1; k += 1) {
StringBuilder sb = new StringBuilder(temp1);
sb.insert(k, c);
result.add(sb.toString());
}
}
}
permutations = result;
//'result' has to be refreshed so that in next run it doesn't contain stale values.
result = new HashSet<String>();
}
}
public static void main(String[] args) {
Set<String> result = permutation("abc");
System.out.println("\nThere are total of " + result.size() + " permutations:");
Iterator<String> it = result.iterator();
while (it.hasNext()) {
System.out.println(it.next());
}
}
}

This one is without recursion
public static void permute(String s) {
if(null==s || s.isEmpty()) {
return;
}
// List containing words formed in each iteration
List<String> strings = new LinkedList<String>();
strings.add(String.valueOf(s.charAt(0))); // add the first element to the list
// Temp list that holds the set of strings for
// appending the current character to all position in each word in the original list
List<String> tempList = new LinkedList<String>();
for(int i=1; i< s.length(); i++) {
for(int j=0; j<strings.size(); j++) {
tempList.addAll(merge(s.charAt(i), strings.get(j)));
}
strings.removeAll(strings);
strings.addAll(tempList);
tempList.removeAll(tempList);
}
for(int i=0; i<strings.size(); i++) {
System.out.println(strings.get(i));
}
}
/**
* helper method that appends the given character at each position in the given string
* and returns a set of such modified strings
* - set removes duplicates if any(in case a character is repeated)
*/
private static Set<String> merge(Character c, String s) {
if(s==null || s.isEmpty()) {
return null;
}
int len = s.length();
StringBuilder sb = new StringBuilder();
Set<String> list = new HashSet<String>();
for(int i=0; i<= len; i++) {
sb = new StringBuilder();
sb.append(s.substring(0, i) + c + s.substring(i, len));
list.add(sb.toString());
}
return list;
}

Well here is an elegant, non-recursive, O(n!) solution:
public static StringBuilder[] permutations(String s) {
if (s.length() == 0)
return null;
int length = fact(s.length());
StringBuilder[] sb = new StringBuilder[length];
for (int i = 0; i < length; i++) {
sb[i] = new StringBuilder();
}
for (int i = 0; i < s.length(); i++) {
char ch = s.charAt(i);
int times = length / (i + 1);
for (int j = 0; j < times; j++) {
for (int k = 0; k < length / times; k++) {
sb[j * length / times + k].insert(k, ch);
}
}
}
return sb;
}

One of the simple solution could be just keep swapping the characters recursively using two pointers.
public static void main(String[] args)
{
String str="abcdefgh";
perm(str);
}
public static void perm(String str)
{ char[] char_arr=str.toCharArray();
helper(char_arr,0);
}
public static void helper(char[] char_arr, int i)
{
if(i==char_arr.length-1)
{
// print the shuffled string
String str="";
for(int j=0; j<char_arr.length; j++)
{
str=str+char_arr[j];
}
System.out.println(str);
}
else
{
for(int j=i; j<char_arr.length; j++)
{
char tmp = char_arr[i];
char_arr[i] = char_arr[j];
char_arr[j] = tmp;
helper(char_arr,i+1);
char tmp1 = char_arr[i];
char_arr[i] = char_arr[j];
char_arr[j] = tmp1;
}
}
}

python implementation
def getPermutation(s, prefix=''):
if len(s) == 0:
print prefix
for i in range(len(s)):
getPermutation(s[0:i]+s[i+1:len(s)],prefix+s[i] )
getPermutation('abcd','')

This is what I did through basic understanding of Permutations and Recursive function calling. Takes a bit of time but it's done independently.
public class LexicographicPermutations {
public static void main(String[] args) {
// TODO Auto-generated method stub
String s="abc";
List<String>combinations=new ArrayList<String>();
combinations=permutations(s);
Collections.sort(combinations);
System.out.println(combinations);
}
private static List<String> permutations(String s) {
// TODO Auto-generated method stub
List<String>combinations=new ArrayList<String>();
if(s.length()==1){
combinations.add(s);
}
else{
for(int i=0;i<s.length();i++){
List<String>temp=permutations(s.substring(0, i)+s.substring(i+1));
for (String string : temp) {
combinations.add(s.charAt(i)+string);
}
}
}
return combinations;
}}
which generates Output as [abc, acb, bac, bca, cab, cba].
Basic logic behind it is
For each character, consider it as 1st character & find the combinations of remaining characters. e.g. [abc](Combination of abc)->.
a->[bc](a x Combination of (bc))->{abc,acb}
b->[ac](b x Combination of (ac))->{bac,bca}
c->[ab](c x Combination of (ab))->{cab,cba}
And then recursively calling each [bc],[ac] & [ab] independently.

Use recursion.
when the input is an empty string the only permutation is an empty string.Try for each of the letters in the string by making it as the first letter and then find all the permutations of the remaining letters using a recursive call.
import java.util.ArrayList;
import java.util.List;
class Permutation {
private static List<String> permutation(String prefix, String str) {
List<String> permutations = new ArrayList<>();
int n = str.length();
if (n == 0) {
permutations.add(prefix);
} else {
for (int i = 0; i < n; i++) {
permutations.addAll(permutation(prefix + str.charAt(i), str.substring(i + 1, n) + str.substring(0, i)));
}
}
return permutations;
}
public static void main(String[] args) {
List<String> perms = permutation("", "abcd");
String[] array = new String[perms.size()];
for (int i = 0; i < perms.size(); i++) {
array[i] = perms.get(i);
}
int x = array.length;
for (final String anArray : array) {
System.out.println(anArray);
}
}
}

this worked for me..
import java.util.Arrays;
public class StringPermutations{
public static void main(String args[]) {
String inputString = "ABC";
permute(inputString.toCharArray(), 0, inputString.length()-1);
}
public static void permute(char[] ary, int startIndex, int endIndex) {
if(startIndex == endIndex){
System.out.println(String.valueOf(ary));
}else{
for(int i=startIndex;i<=endIndex;i++) {
swap(ary, startIndex, i );
permute(ary, startIndex+1, endIndex);
swap(ary, startIndex, i );
}
}
}
public static void swap(char[] ary, int x, int y) {
char temp = ary[x];
ary[x] = ary[y];
ary[y] = temp;
}
}

Java implementation without recursion
public Set<String> permutate(String s){
Queue<String> permutations = new LinkedList<String>();
Set<String> v = new HashSet<String>();
permutations.add(s);
while(permutations.size()!=0){
String str = permutations.poll();
if(!v.contains(str)){
v.add(str);
for(int i = 0;i<str.length();i++){
String c = String.valueOf(str.charAt(i));
permutations.add(str.substring(i+1) + c + str.substring(0,i));
}
}
}
return v;
}

Let me try to tackle this problem with Kotlin:
fun <T> List<T>.permutations(): List<List<T>> {
//escape case
if (this.isEmpty()) return emptyList()
if (this.size == 1) return listOf(this)
if (this.size == 2) return listOf(listOf(this.first(), this.last()), listOf(this.last(), this.first()))
//recursive case
return this.flatMap { lastItem ->
this.minus(lastItem).permutations().map { it.plus(lastItem) }
}
}
Core concept: Break down long list into smaller list + recursion
Long answer with example list [1, 2, 3, 4]:
Even for a list of 4 it already kinda get's confusing trying to list all the possible permutations in your head, and what we need to do is exactly to avoid that. It is easy for us to understand how to make all permutations of list of size 0, 1, and 2, so all we need to do is break them down to any of those sizes and combine them back up correctly. Imagine a jackpot machine: this algorithm will start spinning from the right to the left, and write down
return empty/list of 1 when list size is 0 or 1
handle when list size is 2 (e.g. [3, 4]), and generate the 2 permutations ([3, 4] & [4, 3])
For each item, mark that as the last in the last, and find all the permutations for the rest of the item in the list. (e.g. put [4] on the table, and throw [1, 2, 3] into permutation again)
Now with all permutation it's children, put itself back to the end of the list (e.g.: [1, 2, 3][,4], [1, 3, 2][,4], [2, 3, 1][, 4], ...)

import java.io.IOException;
import java.util.ArrayList;
import java.util.Scanner;
public class hello {
public static void main(String[] args) throws IOException {
hello h = new hello();
h.printcomp();
}
int fact=1;
public void factrec(int a,int k){
if(a>=k)
{fact=fact*k;
k++;
factrec(a,k);
}
else
{System.out.println("The string will have "+fact+" permutations");
}
}
public void printcomp(){
String str;
int k;
Scanner in = new Scanner(System.in);
System.out.println("enter the string whose permutations has to b found");
str=in.next();
k=str.length();
factrec(k,1);
String[] arr =new String[fact];
char[] array = str.toCharArray();
while(p<fact)
printcomprec(k,array,arr);
// if incase u need array containing all the permutation use this
//for(int d=0;d<fact;d++)
//System.out.println(arr[d]);
}
int y=1;
int p = 0;
int g=1;
int z = 0;
public void printcomprec(int k,char array[],String arr[]){
for (int l = 0; l < k; l++) {
for (int b=0;b<k-1;b++){
for (int i=1; i<k-g; i++) {
char temp;
String stri = "";
temp = array[i];
array[i] = array[i + g];
array[i + g] = temp;
for (int j = 0; j < k; j++)
stri += array[j];
arr[z] = stri;
System.out.println(arr[z] + " " + p++);
z++;
}
}
char temp;
temp=array[0];
array[0]=array[y];
array[y]=temp;
if (y >= k-1)
y=y-(k-1);
else
y++;
}
if (g >= k-1)
g=1;
else
g++;
}
}

/** Returns an array list containing all
* permutations of the characters in s. */
public static ArrayList<String> permute(String s) {
ArrayList<String> perms = new ArrayList<>();
int slen = s.length();
if (slen > 0) {
// Add the first character from s to the perms array list.
perms.add(Character.toString(s.charAt(0)));
// Repeat for all additional characters in s.
for (int i = 1; i < slen; ++i) {
// Get the next character from s.
char c = s.charAt(i);
// For each of the strings currently in perms do the following:
int size = perms.size();
for (int j = 0; j < size; ++j) {
// 1. remove the string
String p = perms.remove(0);
int plen = p.length();
// 2. Add plen + 1 new strings to perms. Each new string
// consists of the removed string with the character c
// inserted into it at a unique location.
for (int k = 0; k <= plen; ++k) {
perms.add(p.substring(0, k) + c + p.substring(k));
}
}
}
}
return perms;
}

Here is a straightforward minimalist recursive solution in Java:
public static ArrayList<String> permutations(String s) {
ArrayList<String> out = new ArrayList<String>();
if (s.length() == 1) {
out.add(s);
return out;
}
char first = s.charAt(0);
String rest = s.substring(1);
for (String permutation : permutations(rest)) {
out.addAll(insertAtAllPositions(first, permutation));
}
return out;
}
public static ArrayList<String> insertAtAllPositions(char ch, String s) {
ArrayList<String> out = new ArrayList<String>();
for (int i = 0; i <= s.length(); ++i) {
String inserted = s.substring(0, i) + ch + s.substring(i);
out.add(inserted);
}
return out;
}

We can use factorial to find how many strings started with particular letter.
Example: take the input abcd. (3!) == 6 strings will start with every letter of abcd.
static public int facts(int x){
int sum = 1;
for (int i = 1; i < x; i++) {
sum *= (i+1);
}
return sum;
}
public static void permutation(String str) {
char[] str2 = str.toCharArray();
int n = str2.length;
int permutation = 0;
if (n == 1) {
System.out.println(str2[0]);
} else if (n == 2) {
System.out.println(str2[0] + "" + str2[1]);
System.out.println(str2[1] + "" + str2[0]);
} else {
for (int i = 0; i < n; i++) {
if (true) {
char[] str3 = str.toCharArray();
char temp = str3[i];
str3[i] = str3[0];
str3[0] = temp;
str2 = str3;
}
for (int j = 1, count = 0; count < facts(n-1); j++, count++) {
if (j != n-1) {
char temp1 = str2[j+1];
str2[j+1] = str2[j];
str2[j] = temp1;
} else {
char temp1 = str2[n-1];
str2[n-1] = str2[1];
str2[1] = temp1;
j = 1;
} // end of else block
permutation++;
System.out.print("permutation " + permutation + " is -> ");
for (int k = 0; k < n; k++) {
System.out.print(str2[k]);
} // end of loop k
System.out.println();
} // end of loop j
} // end of loop i
}
}

//insert each character into an arraylist
static ArrayList al = new ArrayList();
private static void findPermutation (String str){
for (int k = 0; k < str.length(); k++) {
addOneChar(str.charAt(k));
}
}
//insert one char into ArrayList
private static void addOneChar(char ch){
String lastPerStr;
String tempStr;
ArrayList locAl = new ArrayList();
for (int i = 0; i < al.size(); i ++ ){
lastPerStr = al.get(i).toString();
//System.out.println("lastPerStr: " + lastPerStr);
for (int j = 0; j <= lastPerStr.length(); j++) {
tempStr = lastPerStr.substring(0,j) + ch +
lastPerStr.substring(j, lastPerStr.length());
locAl.add(tempStr);
//System.out.println("tempStr: " + tempStr);
}
}
if(al.isEmpty()){
al.add(ch);
} else {
al.clear();
al = locAl;
}
}
private static void printArrayList(ArrayList al){
for (int i = 0; i < al.size(); i++) {
System.out.print(al.get(i) + " ");
}
}

//Rotate and create words beginning with all letter possible and push to stack 1
//Read from stack1 and for each word create words with other letters at the next location by rotation and so on
/* eg : man
1. push1 - man, anm, nma
2. pop1 - nma , push2 - nam,nma
pop1 - anm , push2 - amn,anm
pop1 - man , push2 - mna,man
*/
public class StringPermute {
static String str;
static String word;
static int top1 = -1;
static int top2 = -1;
static String[] stringArray1;
static String[] stringArray2;
static int strlength = 0;
public static void main(String[] args) throws IOException {
System.out.println("Enter String : ");
InputStreamReader isr = new InputStreamReader(System.in);
BufferedReader bfr = new BufferedReader(isr);
str = bfr.readLine();
word = str;
strlength = str.length();
int n = 1;
for (int i = 1; i <= strlength; i++) {
n = n * i;
}
stringArray1 = new String[n];
stringArray2 = new String[n];
push(word, 1);
doPermute();
display();
}
public static void push(String word, int x) {
if (x == 1)
stringArray1[++top1] = word;
else
stringArray2[++top2] = word;
}
public static String pop(int x) {
if (x == 1)
return stringArray1[top1--];
else
return stringArray2[top2--];
}
public static void doPermute() {
for (int j = strlength; j >= 2; j--)
popper(j);
}
public static void popper(int length) {
// pop from stack1 , rotate each word n times and push to stack 2
if (top1 > -1) {
while (top1 > -1) {
word = pop(1);
for (int j = 0; j < length; j++) {
rotate(length);
push(word, 2);
}
}
}
// pop from stack2 , rotate each word n times w.r.t position and push to
// stack 1
else {
while (top2 > -1) {
word = pop(2);
for (int j = 0; j < length; j++) {
rotate(length);
push(word, 1);
}
}
}
}
public static void rotate(int position) {
char[] charstring = new char[100];
for (int j = 0; j < word.length(); j++)
charstring[j] = word.charAt(j);
int startpos = strlength - position;
char temp = charstring[startpos];
for (int i = startpos; i < strlength - 1; i++) {
charstring[i] = charstring[i + 1];
}
charstring[strlength - 1] = temp;
word = new String(charstring).trim();
}
public static void display() {
int top;
if (top1 > -1) {
while (top1 > -1)
System.out.println(stringArray1[top1--]);
} else {
while (top2 > -1)
System.out.println(stringArray2[top2--]);
}
}
}

Another simple way is to loop through the string, pick the character that is not used yet and put it to a buffer, continue the loop till the buffer size equals to the string length. I like this back tracking solution better because:
Easy to understand
Easy to avoid duplication
The output is sorted
Here is the java code:
List<String> permute(String str) {
if (str == null) {
return null;
}
char[] chars = str.toCharArray();
boolean[] used = new boolean[chars.length];
List<String> res = new ArrayList<String>();
StringBuilder sb = new StringBuilder();
Arrays.sort(chars);
helper(chars, used, sb, res);
return res;
}
void helper(char[] chars, boolean[] used, StringBuilder sb, List<String> res) {
if (sb.length() == chars.length) {
res.add(sb.toString());
return;
}
for (int i = 0; i < chars.length; i++) {
// avoid duplicates
if (i > 0 && chars[i] == chars[i - 1] && !used[i - 1]) {
continue;
}
// pick the character that has not used yet
if (!used[i]) {
used[i] = true;
sb.append(chars[i]);
helper(chars, used, sb, res);
// back tracking
sb.deleteCharAt(sb.length() - 1);
used[i] = false;
}
}
}
Input str: 1231
Output list: {1123, 1132, 1213, 1231, 1312, 1321, 2113, 2131, 2311, 3112, 3121, 3211}
Noticed that the output is sorted, and there is no duplicate result.

Recursion is not necessary, even you can calculate any permutation directly, this solution uses generics to permute any array.
Here is a good information about this algorihtm.
For C# developers here is more useful implementation.
public static void main(String[] args) {
String word = "12345";
Character[] array = ArrayUtils.toObject(word.toCharArray());
long[] factorials = Permutation.getFactorials(array.length + 1);
for (long i = 0; i < factorials[array.length]; i++) {
Character[] permutation = Permutation.<Character>getPermutation(i, array, factorials);
printPermutation(permutation);
}
}
private static void printPermutation(Character[] permutation) {
for (int i = 0; i < permutation.length; i++) {
System.out.print(permutation[i]);
}
System.out.println();
}
This algorithm has O(N) time and space complexity to calculate each permutation.
public class Permutation {
public static <T> T[] getPermutation(long permutationNumber, T[] array, long[] factorials) {
int[] sequence = generateSequence(permutationNumber, array.length - 1, factorials);
T[] permutation = generatePermutation(array, sequence);
return permutation;
}
public static <T> T[] generatePermutation(T[] array, int[] sequence) {
T[] clone = array.clone();
for (int i = 0; i < clone.length - 1; i++) {
swap(clone, i, i + sequence[i]);
}
return clone;
}
private static int[] generateSequence(long permutationNumber, int size, long[] factorials) {
int[] sequence = new int[size];
for (int j = 0; j < sequence.length; j++) {
long factorial = factorials[sequence.length - j];
sequence[j] = (int) (permutationNumber / factorial);
permutationNumber = (int) (permutationNumber % factorial);
}
return sequence;
}
private static <T> void swap(T[] array, int i, int j) {
T t = array[i];
array[i] = array[j];
array[j] = t;
}
public static long[] getFactorials(int length) {
long[] factorials = new long[length];
long factor = 1;
for (int i = 0; i < length; i++) {
factor *= i <= 1 ? 1 : i;
factorials[i] = factor;
}
return factorials;
}
}

My implementation based on Mark Byers's description above:
static Set<String> permutations(String str){
if (str.isEmpty()){
return Collections.singleton(str);
}else{
Set <String> set = new HashSet<>();
for (int i=0; i<str.length(); i++)
for (String s : permutations(str.substring(0, i) + str.substring(i+1)))
set.add(str.charAt(i) + s);
return set;
}
}

Permutation of String:
public static void main(String args[]) {
permu(0,"ABCD");
}
static void permu(int fixed,String s) {
char[] chr=s.toCharArray();
if(fixed==s.length())
System.out.println(s);
for(int i=fixed;i<s.length();i++) {
char c=chr[i];
chr[i]=chr[fixed];
chr[fixed]=c;
permu(fixed+1,new String(chr));
}
}

Here is another simpler method of doing Permutation of a string.
public class Solution4 {
public static void main(String[] args) {
String a = "Protijayi";
per(a, 0);
}
static void per(String a , int start ) {
//bse case;
if(a.length() == start) {System.out.println(a);}
char[] ca = a.toCharArray();
//swap
for (int i = start; i < ca.length; i++) {
char t = ca[i];
ca[i] = ca[start];
ca[start] = t;
per(new String(ca),start+1);
}
}//per
}

A java implementation to print all the permutations of a given string considering duplicate characters and prints only unique characters is as follow:
import java.util.Set;
import java.util.HashSet;
public class PrintAllPermutations2
{
public static void main(String[] args)
{
String str = "AAC";
PrintAllPermutations2 permutation = new PrintAllPermutations2();
Set<String> uniqueStrings = new HashSet<>();
permutation.permute("", str, uniqueStrings);
}
void permute(String prefixString, String s, Set<String> set)
{
int n = s.length();
if(n == 0)
{
if(!set.contains(prefixString))
{
System.out.println(prefixString);
set.add(prefixString);
}
}
else
{
for(int i=0; i<n; i++)
{
permute(prefixString + s.charAt(i), s.substring(0,i) + s.substring(i+1,n), set);
}
}
}
}

String permutaions using Es6
Using reduce() method
const permutations = str => {
if (str.length <= 2)
return str.length === 2 ? [str, str[1] + str[0]] : [str];
return str
.split('')
.reduce(
(acc, letter, index) =>
acc.concat(permutations(str.slice(0, index) + str.slice(index + 1)).map(val => letter + val)),
[]
);
};
console.log(permutations('STR'));

In case anyone wants to generate the permutations to do something with them, instead of just printing them via a void method:
static List<int[]> permutations(int n) {
class Perm {
private final List<int[]> permutations = new ArrayList<>();
private void perm(int[] array, int step) {
if (step == 1) permutations.add(array.clone());
else for (int i = 0; i < step; i++) {
perm(array, step - 1);
int j = (step % 2 == 0) ? i : 0;
swap(array, step - 1, j);
}
}
private void swap(int[] array, int i, int j) {
int buffer = array[i];
array[i] = array[j];
array[j] = buffer;
}
}
int[] nVector = new int[n];
for (int i = 0; i < n; i++) nVector [i] = i;
Perm perm = new Perm();
perm.perm(nVector, n);
return perm.permutations;
}

Find ascending duplicate pairs in an array

Given an array A with zero index and N integers find equal elements with different positions in the array. Pair of indexes (P,Q) such that 0 <= P < Q < N such that A[P] = A[Q]. My algorithm is below but I am looking for a O(N*logN) solution.
public int solution(int[] A)
{
int N = A.Length;
int count = 0;
for (int j = 0; j < N; j++)
{
count += FindPairs(A[j], j, A);
}
return count;
}
public int FindPairs(int item, int ci, int[] A)
{
int len = A.Length;
int counter=0;
int k = ci+1;
while (k < len)
{
if (item == A[k])
counter++;
k++;
}
return counter;
}

From your code, it looks like the goal is to return the count of ascending duplicate pairs in A.
We observe that if there are m occurrences of the number x in A, then the number of ascending duplicate pairs of the value x is m choose 2, or m (m - 1) / 2.
So, we sum up m (m - 1) / 2 for each unique x, giving us the answer.
In pseudocode, this looks like:
count = new Dictionary();
foreach a in A {
count[a]++;
}
total = 0;
foreach key, value in count {
total += value * (value - 1) / 2;
}
return total;
This algorithm is O(N).

Recent interview question … here is what I did:
using System;
using System.Collections.Generic;
using System.Linq;
namespace Codility
{
internal class Program
{
public struct Indice
{
public Indice(int p, int q)
{
P = p;
Q = q;
}
public int P;
public int Q;
public override string ToString()
{
return string.Format("({0}, {1})", P, Q);
}
}
private static void Main(string[] args)
{
// 0 1 2 3 4 5
int[] list = new int[] {3,3,3,3,3,3};
int answer = GetPairCount(list);
Console.WriteLine("answer = " + answer);
Console.ReadLine();
}
private static int GetPairCount(int[] A)
{
if (A.Length < 2) return 0;
Dictionary<int, Dictionary<Indice, Indice>> tracker = new Dictionary<int, Dictionary<Indice, Indice>>();
for (int i = 0; i < A.Length; i++)
{
int val = A[i];
if (!tracker.ContainsKey(val))
{
Dictionary<Indice, Indice> list = new Dictionary<Indice, Indice>();
Indice seed = new Indice(i, -1);
list.Add(seed, seed);
tracker.Add(val, list);
}
else
{
Dictionary<Indice, Indice> list = tracker[val];
foreach (KeyValuePair<Indice,Indice> item in list.ToList())
{
Indice left = new Indice(item.Value.P, i);
Indice right = new Indice(i, item.Value.Q);
if (!list.ContainsKey(left))
{
list.Add(left, left);
Console.WriteLine("left= " + left);
}
if (!list.ContainsKey(right))
{
list.Add(right, right);
Console.WriteLine("\t\tright= " + right);
}
}
}
}
return tracker.SelectMany(kvp => kvp.Value).Count(num => num.Value.Q > num.Value.P);
}
}
}

I think this is best version I got in c#.
static void Main(string[] args)
{
var a = new int[6] { 3, 5, 6, 3, 3, 5 };
//Push the indices into an array:
int[] indices = new int[a.Count()];
for (int p = 0; p < a.Count(); ++p) indices[p] = p;
//Sort the indices according to the value of the corresponding element in a:
Array.Sort(indices, (k, l) =>Compare(a[k], a[l]));
//Then just pull out blocks of indices with equal corresponding elements from indices:
int count = 0;
int i = 0;
while (i < indices.Count())
{
int start = i;
while (i < indices.Count() && a[indices[i]] == a[indices[start]])
{
++i;
}
int thisCount = i - start;
int numPairs = thisCount * (thisCount - 1) / 2;
count += numPairs;
}
Console.WriteLine(count);
Console.ReadKey();
}
//Compare function to return interger
private static int Compare(int v1, int v2)
{
if (v2 > v1)
return 1;
if (v1 == v2)
return 0;
else
return -1;
}
This approach has O(n log n) complexity overall, because of the sorting. The counting of the groups is linear.

Try this:
private static int GetIdenticalPairCount(int[] input)
{
int identicalPairCount = 0;
Dictionary<int, int> identicalCountMap = new Dictionary<int, int>();
foreach (int i in input)
{
if (identicalCountMap.ContainsKey(i))
{
identicalCountMap[i] = identicalCountMap[i] + 1;
if (identicalCountMap[i] > 1)
{
identicalPairCount += identicalCountMap[i];
}
else
{
identicalPairCount++;
}
}
else
{
identicalCountMap.Add(i, 0);
}
}
return identicalPairCount;
}

Test my version:
public int solution(int[] A)
{
int N = A.Length;
int count = 0;
for (int j = 0; j < N - 1; j++)
for (int i = j + 1; i < N; i++)
if (A[i] == A[j])
count++;
return count;
}

permute numbers without repeating [duplicate]

A common task in programming interviews (not from my experience of interviews though) is to take a string or an integer and list every possible permutation.
Is there an example of how this is done and the logic behind solving such a problem?
I've seen a few code snippets but they weren't well commented/explained and thus hard to follow.

First of all: it smells like recursion of course!
Since you also wanted to know the principle, I did my best to explain it human language. I think recursion is very easy most of the times. You only have to grasp two steps:
The first step
All the other steps (all with the same logic)
In human language:
In short:
The permutation of 1 element is one element.
The permutation of a set of elements is a list each of the elements, concatenated with every permutation of the other elements.
Example:
If the set just has one element -->
return it.
perm(a) -> a
If the set has two characters: for
each element in it: return the
element, with the permutation of the
rest of the elements added, like so:
perm(ab) ->
a + perm(b) -> ab
b + perm(a) -> ba
Further: for each character in the set: return a character, concatenated with a permutation of > the rest of the set
perm(abc) ->
a + perm(bc) --> abc, acb
b + perm(ac) --> bac, bca
c + perm(ab) --> cab, cba
perm(abc...z) -->
a + perm(...), b + perm(....)
....
I found the pseudocode on http://www.programmersheaven.com/mb/Algorithms/369713/369713/permutation-algorithm-help/:
makePermutations(permutation) {
if (length permutation < required length) {
for (i = min digit to max digit) {
if (i not in permutation) {
makePermutations(permutation+i)
}
}
}
else {
add permutation to list
}
}
C#
OK, and something more elaborate (and since it is tagged c #), from http://radio.weblogs.com/0111551/stories/2002/10/14/permutations.html :
Rather lengthy, but I decided to copy it anyway, so the post is not dependent on the original.
The function takes a string of characters, and writes down every possible permutation of that exact string, so for example, if "ABC" has been supplied, should spill out:
ABC, ACB, BAC, BCA, CAB, CBA.
Code:
class Program
{
private static void Swap(ref char a, ref char b)
{
if (a == b) return;
var temp = a;
a = b;
b = temp;
}
public static void GetPer(char[] list)
{
int x = list.Length - 1;
GetPer(list, 0, x);
}
private static void GetPer(char[] list, int k, int m)
{
if (k == m)
{
Console.Write(list);
}
else
for (int i = k; i <= m; i++)
{
Swap(ref list[k], ref list[i]);
GetPer(list, k + 1, m);
Swap(ref list[k], ref list[i]);
}
}
static void Main()
{
string str = "sagiv";
char[] arr = str.ToCharArray();
GetPer(arr);
}
}

It's just two lines of code if LINQ is allowed to use. Please see my answer here.
EDIT
Here is my generic function which can return all the permutations (not combinations) from a list of T:
static IEnumerable<IEnumerable<T>>
GetPermutations<T>(IEnumerable<T> list, int length)
{
if (length == 1) return list.Select(t => new T[] { t });
return GetPermutations(list, length - 1)
.SelectMany(t => list.Where(e => !t.Contains(e)),
(t1, t2) => t1.Concat(new T[] { t2 }));
}
Example:
IEnumerable<IEnumerable<int>> result =
GetPermutations(Enumerable.Range(1, 3), 3);
Output - a list of integer-lists:
{1,2,3} {1,3,2} {2,1,3} {2,3,1} {3,1,2} {3,2,1}
As this function uses LINQ so it requires .net 3.5 or higher.

Here I have found the solution. It was written in Java, but I have converted it to C#. I hope it will help you.
Here's the code in C#:
static void Main(string[] args)
{
string str = "ABC";
char[] charArry = str.ToCharArray();
Permute(charArry, 0, 2);
Console.ReadKey();
}
static void Permute(char[] arry, int i, int n)
{
int j;
if (i==n)
Console.WriteLine(arry);
else
{
for(j = i; j <=n; j++)
{
Swap(ref arry[i],ref arry[j]);
Permute(arry,i+1,n);
Swap(ref arry[i], ref arry[j]); //backtrack
}
}
}
static void Swap(ref char a, ref char b)
{
char tmp;
tmp = a;
a=b;
b = tmp;
}

Recursion is not necessary, here is good information about this solution.
var values1 = new[] { 1, 2, 3, 4, 5 };
foreach (var permutation in values1.GetPermutations())
{
Console.WriteLine(string.Join(", ", permutation));
}
var values2 = new[] { 'a', 'b', 'c', 'd', 'e' };
foreach (var permutation in values2.GetPermutations())
{
Console.WriteLine(string.Join(", ", permutation));
}
Console.ReadLine();
I have been used this algorithm for years, it has O(N) time and space complexity to calculate each permutation.
public static class SomeExtensions
{
public static IEnumerable<IEnumerable<T>> GetPermutations<T>(this IEnumerable<T> enumerable)
{
var array = enumerable as T[] ?? enumerable.ToArray();
var factorials = Enumerable.Range(0, array.Length + 1)
.Select(Factorial)
.ToArray();
for (var i = 0L; i < factorials[array.Length]; i++)
{
var sequence = GenerateSequence(i, array.Length - 1, factorials);
yield return GeneratePermutation(array, sequence);
}
}
private static IEnumerable<T> GeneratePermutation<T>(T[] array, IReadOnlyList<int> sequence)
{
var clone = (T[]) array.Clone();
for (int i = 0; i < clone.Length - 1; i++)
{
Swap(ref clone[i], ref clone[i + sequence[i]]);
}
return clone;
}
private static int[] GenerateSequence(long number, int size, IReadOnlyList<long> factorials)
{
var sequence = new int[size];
for (var j = 0; j < sequence.Length; j++)
{
var facto = factorials[sequence.Length - j];
sequence[j] = (int)(number / facto);
number = (int)(number % facto);
}
return sequence;
}
static void Swap<T>(ref T a, ref T b)
{
T temp = a;
a = b;
b = temp;
}
private static long Factorial(int n)
{
long result = n;
for (int i = 1; i < n; i++)
{
result = result * i;
}
return result;
}
}

class Program
{
public static void Main(string[] args)
{
Permutation("abc");
}
static void Permutation(string rest, string prefix = "")
{
if (string.IsNullOrEmpty(rest)) Console.WriteLine(prefix);
// Each letter has a chance to be permutated
for (int i = 0; i < rest.Length; i++)
{
Permutation(rest.Remove(i, 1), prefix + rest[i]);
}
}
}

Slightly modified version in C# that yields needed permutations in an array of ANY type.
// USAGE: create an array of any type, and call Permutations()
var vals = new[] {"a", "bb", "ccc"};
foreach (var v in Permutations(vals))
Console.WriteLine(string.Join(",", v)); // Print values separated by comma
public static IEnumerable<T[]> Permutations<T>(T[] values, int fromInd = 0)
{
if (fromInd + 1 == values.Length)
yield return values;
else
{
foreach (var v in Permutations(values, fromInd + 1))
yield return v;
for (var i = fromInd + 1; i < values.Length; i++)
{
SwapValues(values, fromInd, i);
foreach (var v in Permutations(values, fromInd + 1))
yield return v;
SwapValues(values, fromInd, i);
}
}
}
private static void SwapValues<T>(T[] values, int pos1, int pos2)
{
if (pos1 != pos2)
{
T tmp = values[pos1];
values[pos1] = values[pos2];
values[pos2] = tmp;
}
}

First of all, sets have permutations, not strings or integers, so I'll just assume you mean "the set of characters in a string."
Note that a set of size n has n! n-permutations.
The following pseudocode (from Wikipedia), called with k = 1...n! will give all the permutations:
function permutation(k, s) {
for j = 2 to length(s) {
swap s[(k mod j) + 1] with s[j]; // note that our array is indexed starting at 1
k := k / j; // integer division cuts off the remainder
}
return s;
}
Here's the equivalent Python code (for 0-based array indexes):
def permutation(k, s):
r = s[:]
for j in range(2, len(s)+1):
r[j-1], r[k%j] = r[k%j], r[j-1]
k = k/j+1
return r

I liked FBryant87 approach since it's simple. Unfortunately, it does like many other "solutions" not offer all permutations or of e.g. an integer if it contains the same digit more than once. Take 656123 as an example. The line:
var tail = chars.Except(new List<char>(){c});
using Except will cause all occurrences to be removed, i.e. when c = 6, two digits are removed and we are left with e.g. 5123. Since none of the solutions I tried solved this, I decided to try and solve it myself by FBryant87's code as base. This is what I came up with:
private static List<string> FindPermutations(string set)
{
var output = new List<string>();
if (set.Length == 1)
{
output.Add(set);
}
else
{
foreach (var c in set)
{
// Remove one occurrence of the char (not all)
var tail = set.Remove(set.IndexOf(c), 1);
foreach (var tailPerms in FindPermutations(tail))
{
output.Add(c + tailPerms);
}
}
}
return output;
}
I simply just remove the first found occurrence using .Remove and .IndexOf. Seems to work as intended for my usage at least. I'm sure it could be made cleverer.
One thing to note though: The resulting list may contain duplicates, so make sure you either make the method return e.g. a HashSet instead or remove the duplicates after the return using any method you like.

Here is a simple solution in c# using recursion,
void Main()
{
string word = "abc";
WordPermuatation("",word);
}
void WordPermuatation(string prefix, string word)
{
int n = word.Length;
if (n == 0) { Console.WriteLine(prefix); }
else
{
for (int i = 0; i < n; i++)
{
WordPermuatation(prefix + word[i],word.Substring(0, i) + word.Substring(i + 1, n - (i+1)));
}
}
}

Here's a purely functional F# implementation:
let factorial i =
let rec fact n x =
match n with
| 0 -> 1
| 1 -> x
| _ -> fact (n-1) (x*n)
fact i 1
let swap (arr:'a array) i j = [| for k in 0..(arr.Length-1) -> if k = i then arr.[j] elif k = j then arr.[i] else arr.[k] |]
let rec permutation (k:int,j:int) (r:'a array) =
if j = (r.Length + 1) then r
else permutation (k/j+1, j+1) (swap r (j-1) (k%j))
let permutations (source:'a array) = seq { for k = 0 to (source |> Array.length |> factorial) - 1 do yield permutation (k,2) source }
Performance can be greatly improved by changing swap to take advantage of the mutable nature of CLR arrays, but this implementation is thread safe with regards to the source array and that may be desirable in some contexts.
Also, for arrays with more than 16 elements int must be replaced with types with greater/arbitrary precision as factorial 17 results in an int32 overflow.

Here is an easy to understand permutaion function for both string and integer as input. With this you can even set your output length(which in normal case it is equal to input length)
String
static ICollection<string> result;
public static ICollection<string> GetAllPermutations(string str, int outputLength)
{
result = new List<string>();
MakePermutations(str.ToCharArray(), string.Empty, outputLength);
return result;
}
private static void MakePermutations(
char[] possibleArray,//all chars extracted from input
string permutation,
int outputLength//the length of output)
{
if (permutation.Length < outputLength)
{
for (int i = 0; i < possibleArray.Length; i++)
{
var tempList = possibleArray.ToList<char>();
tempList.RemoveAt(i);
MakePermutations(tempList.ToArray(),
string.Concat(permutation, possibleArray[i]), outputLength);
}
}
else if (!result.Contains(permutation))
result.Add(permutation);
}
and for Integer just change the caller method and MakePermutations() remains untouched:
public static ICollection<int> GetAllPermutations(int input, int outputLength)
{
result = new List<string>();
MakePermutations(input.ToString().ToCharArray(), string.Empty, outputLength);
return result.Select(m => int.Parse(m)).ToList<int>();
}
example 1: GetAllPermutations("abc",3);
"abc" "acb" "bac" "bca" "cab" "cba"
example 2: GetAllPermutations("abcd",2);
"ab" "ac" "ad" "ba" "bc" "bd" "ca" "cb" "cd" "da" "db" "dc"
example 3: GetAllPermutations(486,2);
48 46 84 86 64 68

Building on #Peter's solution, here's a version that declares a simple LINQ-style Permutations() extension method that works on any IEnumerable<T>.
Usage (on string characters example):
foreach (var permutation in "abc".Permutations())
{
Console.WriteLine(string.Join(", ", permutation));
}
Outputs:
a, b, c
a, c, b
b, a, c
b, c, a
c, b, a
c, a, b
Or on any other collection type:
foreach (var permutation in (new[] { "Apples", "Oranges", "Pears"}).Permutations())
{
Console.WriteLine(string.Join(", ", permutation));
}
Outputs:
Apples, Oranges, Pears
Apples, Pears, Oranges
Oranges, Apples, Pears
Oranges, Pears, Apples
Pears, Oranges, Apples
Pears, Apples, Oranges
using System;
using System.Collections.Generic;
using System.Linq;
public static class PermutationExtension
{
public static IEnumerable<T[]> Permutations<T>(this IEnumerable<T> source)
{
var sourceArray = source.ToArray();
var results = new List<T[]>();
Permute(sourceArray, 0, sourceArray.Length - 1, results);
return results;
}
private static void Swap<T>(ref T a, ref T b)
{
T tmp = a;
a = b;
b = tmp;
}
private static void Permute<T>(T[] elements, int recursionDepth, int maxDepth, ICollection<T[]> results)
{
if (recursionDepth == maxDepth)
{
results.Add(elements.ToArray());
return;
}
for (var i = recursionDepth; i <= maxDepth; i++)
{
Swap(ref elements[recursionDepth], ref elements[i]);
Permute(elements, recursionDepth + 1, maxDepth, results);
Swap(ref elements[recursionDepth], ref elements[i]);
}
}
}

Here is the function which will print all permutaion.
This function implements logic Explained by peter.
public class Permutation
{
//http://www.java2s.com/Tutorial/Java/0100__Class-Definition/RecursivemethodtofindallpermutationsofaString.htm
public static void permuteString(String beginningString, String endingString)
{
if (endingString.Length <= 1)
Console.WriteLine(beginningString + endingString);
else
for (int i = 0; i < endingString.Length; i++)
{
String newString = endingString.Substring(0, i) + endingString.Substring(i + 1);
permuteString(beginningString + endingString.ElementAt(i), newString);
}
}
}
static void Main(string[] args)
{
Permutation.permuteString(String.Empty, "abc");
Console.ReadLine();
}

The below is my implementation of permutation . Don't mind the variable names, as i was doing it for fun :)
class combinations
{
static void Main()
{
string choice = "y";
do
{
try
{
Console.WriteLine("Enter word :");
string abc = Console.ReadLine().ToString();
Console.WriteLine("Combinatins for word :");
List<string> final = comb(abc);
int count = 1;
foreach (string s in final)
{
Console.WriteLine("{0} --> {1}", count++, s);
}
Console.WriteLine("Do you wish to continue(y/n)?");
choice = Console.ReadLine().ToString();
}
catch (Exception exc)
{
Console.WriteLine(exc);
}
} while (choice == "y" || choice == "Y");
}
static string swap(string test)
{
return swap(0, 1, test);
}
static List<string> comb(string test)
{
List<string> sec = new List<string>();
List<string> first = new List<string>();
if (test.Length == 1) first.Add(test);
else if (test.Length == 2) { first.Add(test); first.Add(swap(test)); }
else if (test.Length > 2)
{
sec = generateWords(test);
foreach (string s in sec)
{
string init = s.Substring(0, 1);
string restOfbody = s.Substring(1, s.Length - 1);
List<string> third = comb(restOfbody);
foreach (string s1 in third)
{
if (!first.Contains(init + s1)) first.Add(init + s1);
}
}
}
return first;
}
static string ShiftBack(string abc)
{
char[] arr = abc.ToCharArray();
char temp = arr[0];
string wrd = string.Empty;
for (int i = 1; i < arr.Length; i++)
{
wrd += arr[i];
}
wrd += temp;
return wrd;
}
static List<string> generateWords(string test)
{
List<string> final = new List<string>();
if (test.Length == 1)
final.Add(test);
else
{
final.Add(test);
string holdString = test;
while (final.Count < test.Length)
{
holdString = ShiftBack(holdString);
final.Add(holdString);
}
}
return final;
}
static string swap(int currentPosition, int targetPosition, string temp)
{
char[] arr = temp.ToCharArray();
char t = arr[currentPosition];
arr[currentPosition] = arr[targetPosition];
arr[targetPosition] = t;
string word = string.Empty;
for (int i = 0; i < arr.Length; i++)
{
word += arr[i];
}
return word;
}
}

Here's a high level example I wrote which illustrates the human language explanation Peter gave:
public List<string> FindPermutations(string input)
{
if (input.Length == 1)
return new List<string> { input };
var perms = new List<string>();
foreach (var c in input)
{
var others = input.Remove(input.IndexOf(c), 1);
perms.AddRange(FindPermutations(others).Select(perm => c + perm));
}
return perms;
}

This is my solution which it is easy for me to understand
class ClassicPermutationProblem
{
ClassicPermutationProblem() { }
private static void PopulatePosition<T>(List<List<T>> finalList, List<T> list, List<T> temp, int position)
{
foreach (T element in list)
{
List<T> currentTemp = temp.ToList();
if (!currentTemp.Contains(element))
currentTemp.Add(element);
else
continue;
if (position == list.Count)
finalList.Add(currentTemp);
else
PopulatePosition(finalList, list, currentTemp, position + 1);
}
}
public static List<List<int>> GetPermutations(List<int> list)
{
List<List<int>> results = new List<List<int>>();
PopulatePosition(results, list, new List<int>(), 1);
return results;
}
}
static void Main(string[] args)
{
List<List<int>> results = ClassicPermutationProblem.GetPermutations(new List<int>() { 1, 2, 3 });
}

If performance and memory is an issue, I suggest this very efficient implementation. According to Heap's algorithm in Wikipedia, it should be the fastest. Hope it will fits your need :-) !
Just as comparison of this with a Linq implementation for 10! (code included):
This: 36288000 items in 235 millisecs
Linq: 36288000 items in 50051 millisecs
using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;
using System.Runtime.CompilerServices;
using System.Text;
namespace WpfPermutations
{
/// <summary>
/// EO: 2016-04-14
/// Generator of all permutations of an array of anything.
/// Base on Heap's Algorithm. See: https://en.wikipedia.org/wiki/Heap%27s_algorithm#cite_note-3
/// </summary>
public static class Permutations
{
/// <summary>
/// Heap's algorithm to find all pmermutations. Non recursive, more efficient.
/// </summary>
/// <param name="items">Items to permute in each possible ways</param>
/// <param name="funcExecuteAndTellIfShouldStop"></param>
/// <returns>Return true if cancelled</returns>
public static bool ForAllPermutation<T>(T[] items, Func<T[], bool> funcExecuteAndTellIfShouldStop)
{
int countOfItem = items.Length;
if (countOfItem <= 1)
{
return funcExecuteAndTellIfShouldStop(items);
}
var indexes = new int[countOfItem];
for (int i = 0; i < countOfItem; i++)
{
indexes[i] = 0;
}
if (funcExecuteAndTellIfShouldStop(items))
{
return true;
}
for (int i = 1; i < countOfItem;)
{
if (indexes[i] < i)
{ // On the web there is an implementation with a multiplication which should be less efficient.
if ((i & 1) == 1) // if (i % 2 == 1) ... more efficient ??? At least the same.
{
Swap(ref items[i], ref items[indexes[i]]);
}
else
{
Swap(ref items[i], ref items[0]);
}
if (funcExecuteAndTellIfShouldStop(items))
{
return true;
}
indexes[i]++;
i = 1;
}
else
{
indexes[i++] = 0;
}
}
return false;
}
/// <summary>
/// This function is to show a linq way but is far less efficient
/// </summary>
/// <typeparam name="T"></typeparam>
/// <param name="list"></param>
/// <param name="length"></param>
/// <returns></returns>
static IEnumerable<IEnumerable<T>> GetPermutations<T>(IEnumerable<T> list, int length)
{
if (length == 1) return list.Select(t => new T[] { t });
return GetPermutations(list, length - 1)
.SelectMany(t => list.Where(e => !t.Contains(e)),
(t1, t2) => t1.Concat(new T[] { t2 }));
}
/// <summary>
/// Swap 2 elements of same type
/// </summary>
/// <typeparam name="T"></typeparam>
/// <param name="a"></param>
/// <param name="b"></param>
[MethodImpl(MethodImplOptions.AggressiveInlining)]
static void Swap<T>(ref T a, ref T b)
{
T temp = a;
a = b;
b = temp;
}
/// <summary>
/// Func to show how to call. It does a little test for an array of 4 items.
/// </summary>
public static void Test()
{
ForAllPermutation("123".ToCharArray(), (vals) =>
{
Debug.Print(String.Join("", vals));
return false;
});
int[] values = new int[] { 0, 1, 2, 4 };
Debug.Print("Non Linq");
ForAllPermutation(values, (vals) =>
{
Debug.Print(String.Join("", vals));
return false;
});
Debug.Print("Linq");
foreach(var v in GetPermutations(values, values.Length))
{
Debug.Print(String.Join("", v));
}
// Performance
int count = 0;
values = new int[10];
for(int n = 0; n < values.Length; n++)
{
values[n] = n;
}
Stopwatch stopWatch = new Stopwatch();
stopWatch.Reset();
stopWatch.Start();
ForAllPermutation(values, (vals) =>
{
foreach(var v in vals)
{
count++;
}
return false;
});
stopWatch.Stop();
Debug.Print($"Non Linq {count} items in {stopWatch.ElapsedMilliseconds} millisecs");
count = 0;
stopWatch.Reset();
stopWatch.Start();
foreach (var vals in GetPermutations(values, values.Length))
{
foreach (var v in vals)
{
count++;
}
}
stopWatch.Stop();
Debug.Print($"Linq {count} items in {stopWatch.ElapsedMilliseconds} millisecs");
}
}
}

Here's my solution in JavaScript (NodeJS). The main idea is that we take one element at a time, "remove it" from the string, vary the rest of the characters, and insert the element at the front.
function perms (string) {
if (string.length == 0) {
return [];
}
if (string.length == 1) {
return [string];
}
var list = [];
for(var i = 0; i < string.length; i++) {
var invariant = string[i];
var rest = string.substr(0, i) + string.substr(i + 1);
var newPerms = perms(rest);
for (var j = 0; j < newPerms.length; j++) {
list.push(invariant + newPerms[j]);
}
}
return list;
}
module.exports = perms;
And here are the tests:
require('should');
var permutations = require('../src/perms');
describe('permutations', function () {
it('should permute ""', function () {
permutations('').should.eql([]);
})
it('should permute "1"', function () {
permutations('1').should.eql(['1']);
})
it('should permute "12"', function () {
permutations('12').should.eql(['12', '21']);
})
it('should permute "123"', function () {
var expected = ['123', '132', '321', '213', '231', '312'];
var actual = permutations('123');
expected.forEach(function (e) {
actual.should.containEql(e);
})
})
it('should permute "1234"', function () {
// Wolfram Alpha FTW!
var expected = ['1234', '1243', '1324', '1342', '1423', '1432', '2134', '2143', '2314', '2341', '2413', '2431', '3124', '3142', '3214', '3241', '3412', '3421', '4123', '4132'];
var actual = permutations('1234');
expected.forEach(function (e) {
actual.should.containEql(e);
})
})
})

Here is the simplest solution I can think of:
let rec distribute e = function
| [] -> [[e]]
| x::xs' as xs -> (e::xs)::[for xs in distribute e xs' -> x::xs]
let permute xs = Seq.fold (fun ps x -> List.collect (distribute x) ps) [[]] xs
The distribute function takes a new element e and an n-element list and returns a list of n+1 lists each of which has e inserted at a different place. For example, inserting 10 at each of the four possible places in the list [1;2;3]:
> distribute 10 [1..3];;
val it : int list list =
[[10; 1; 2; 3]; [1; 10; 2; 3]; [1; 2; 10; 3]; [1; 2; 3; 10]]
The permute function folds over each element in turn distributing over the permutations accumulated so far, culminating in all permutations. For example, the 6 permutations of the list [1;2;3]:
> permute [1;2;3];;
val it : int list list =
[[3; 2; 1]; [2; 3; 1]; [2; 1; 3]; [3; 1; 2]; [1; 3; 2]; [1; 2; 3]]
Changing the fold to a scan in order to keep the intermediate accumulators sheds some light on how the permutations are generated an element at a time:
> Seq.scan (fun ps x -> List.collect (distribute x) ps) [[]] [1..3];;
val it : seq<int list list> =
seq
[[[]]; [[1]]; [[2; 1]; [1; 2]];
[[3; 2; 1]; [2; 3; 1]; [2; 1; 3]; [3; 1; 2]; [1; 3; 2]; [1; 2; 3]]]

Lists permutations of a string. Avoids duplication when characters are repeated:
using System;
using System.Collections;
class Permutation{
static IEnumerable Permutations(string word){
if (word == null || word.Length <= 1) {
yield return word;
yield break;
}
char firstChar = word[0];
foreach( string subPermute in Permutations (word.Substring (1)) ) {
int indexOfFirstChar = subPermute.IndexOf (firstChar);
if (indexOfFirstChar == -1) indexOfFirstChar = subPermute.Length;
for( int index = 0; index <= indexOfFirstChar; index++ )
yield return subPermute.Insert (index, new string (firstChar, 1));
}
}
static void Main(){
foreach( var permutation in Permutations ("aab") )
Console.WriteLine (permutation);
}
}

//Generic C# Method
private static List<T[]> GetPerms<T>(T[] input, int startIndex = 0)
{
var perms = new List<T[]>();
var l = input.Length - 1;
if (l == startIndex)
perms.Add(input);
else
{
for (int i = startIndex; i <= l; i++)
{
var copy = input.ToArray(); //make copy
var temp = copy[startIndex];
copy[startIndex] = copy[i];
copy[i] = temp;
perms.AddRange(GetPerms(copy, startIndex + 1));
}
}
return perms;
}
//usages
char[] charArray = new char[] { 'A', 'B', 'C' };
var charPerms = GetPerms(charArray);
string[] stringArray = new string[] { "Orange", "Mango", "Apple" };
var stringPerms = GetPerms(stringArray);
int[] intArray = new int[] { 1, 2, 3 };
var intPerms = GetPerms(intArray);

Base/Revise on Pengyang answer
And inspired from permutations-in-javascript
The c# version FunctionalPermutations should be this
static IEnumerable<IEnumerable<T>> FunctionalPermutations<T>(IEnumerable<T> elements, int length)
{
if (length < 2) return elements.Select(t => new T[] { t });
/* Pengyang answser..
return _recur_(list, length - 1).SelectMany(t => list.Where(e => !t.Contains(e)),(t1, t2) => t1.Concat(new T[] { t2 }));
*/
return elements.SelectMany((element_i, i) =>
FunctionalPermutations(elements.Take(i).Concat(elements.Skip(i + 1)), length - 1)
.Select(sub_ei => new[] { element_i }.Concat(sub_ei)));
}

I hope this will suffice:
using System;
public class Program
{
public static void Main()
{
//Example using word cat
permute("cat");
}
static void permute(string word){
for(int i=0; i < word.Length; i++){
char start = word[0];
for(int j=1; j < word.Length; j++){
string left = word.Substring(1,j-1);
string right = word.Substring(j);
Console.WriteLine(start+right+left);
}
if(i+1 < word.Length){
word = wordChange(word, i + 1);
}
}
}
static string wordChange(string word, int index){
string newWord = "";
for(int i=0; i<word.Length; i++){
if(i== 0)
newWord += word[index];
else if(i== index)
newWord += word[0];
else
newWord += word[i];
}
return newWord;
}
output:
cat
cta
act
atc
tca
tac

Here is the function which will print all permutations recursively.
public void Permutations(string input, StringBuilder sb)
{
if (sb.Length == input.Length)
{
Console.WriteLine(sb.ToString());
return;
}
char[] inChar = input.ToCharArray();
for (int i = 0; i < input.Length; i++)
{
if (!sb.ToString().Contains(inChar[i]))
{
sb.Append(inChar[i]);
Permutations(input, sb);
RemoveChar(sb, inChar[i]);
}
}
}
private bool RemoveChar(StringBuilder input, char toRemove)
{
int index = input.ToString().IndexOf(toRemove);
if (index >= 0)
{
input.Remove(index, 1);
return true;
}
return false;
}

class Permutation
{
public static List<string> Permutate(string seed, List<string> lstsList)
{
loopCounter = 0;
// string s="\w{0,2}";
var lstStrs = PermuateRecursive(seed);
Trace.WriteLine("Loop counter :" + loopCounter);
return lstStrs;
}
// Recursive function to find permutation
private static List<string> PermuateRecursive(string seed)
{
List<string> lstStrs = new List<string>();
if (seed.Length > 2)
{
for (int i = 0; i < seed.Length; i++)
{
str = Swap(seed, 0, i);
PermuateRecursive(str.Substring(1, str.Length - 1)).ForEach(
s =>
{
lstStrs.Add(str[0] + s);
loopCounter++;
});
;
}
}
else
{
lstStrs.Add(seed);
lstStrs.Add(Swap(seed, 0, 1));
}
return lstStrs;
}
//Loop counter variable to count total number of loop execution in various functions
private static int loopCounter = 0;
//Non recursive version of permuation function
public static List<string> Permutate(string seed)
{
loopCounter = 0;
List<string> strList = new List<string>();
strList.Add(seed);
for (int i = 0; i < seed.Length; i++)
{
int count = strList.Count;
for (int j = i + 1; j < seed.Length; j++)
{
for (int k = 0; k < count; k++)
{
strList.Add(Swap(strList[k], i, j));
loopCounter++;
}
}
}
Trace.WriteLine("Loop counter :" + loopCounter);
return strList;
}
private static string Swap(string seed, int p, int p2)
{
Char[] chars = seed.ToCharArray();
char temp = chars[p2];
chars[p2] = chars[p];
chars[p] = temp;
return new string(chars);
}
}

Here is a C# answer which is a little simplified.
public static void StringPermutationsDemo()
{
strBldr = new StringBuilder();
string result = Permute("ABCD".ToCharArray(), 0);
MessageBox.Show(result);
}
static string Permute(char[] elementsList, int startIndex)
{
if (startIndex == elementsList.Length)
{
foreach (char element in elementsList)
{
strBldr.Append(" " + element);
}
strBldr.AppendLine("");
}
else
{
for (int tempIndex = startIndex; tempIndex <= elementsList.Length - 1; tempIndex++)
{
Swap(ref elementsList[startIndex], ref elementsList[tempIndex]);
Permute(elementsList, (startIndex + 1));
Swap(ref elementsList[startIndex], ref elementsList[tempIndex]);
}
}
return strBldr.ToString();
}
static void Swap(ref char Char1, ref char Char2)
{
char tempElement = Char1;
Char1 = Char2;
Char2 = tempElement;
}
Output:
1 2 3
1 3 2
2 1 3
2 3 1
3 2 1
3 1 2

Here is one more implementation of the algo mentioned.
public class Program
{
public static void Main(string[] args)
{
string str = "abcefgh";
var astr = new Permutation().GenerateFor(str);
Console.WriteLine(astr.Length);
foreach(var a in astr)
{
Console.WriteLine(a);
}
//a.ForEach(Console.WriteLine);
}
}
class Permutation
{
public string[] GenerateFor(string s)
{
if(s.Length == 1)
{
return new []{s};
}
else if(s.Length == 2)
{
return new []{s[1].ToString()+s[0].ToString(),s[0].ToString()+s[1].ToString()};
}
var comb = new List<string>();
foreach(var c in s)
{
string cStr = c.ToString();
var sToProcess = s.Replace(cStr,"");
if (!string.IsNullOrEmpty(sToProcess) && sToProcess.Length>0)
{
var conCatStr = GenerateFor(sToProcess);
foreach(var a in conCatStr)
{
comb.Add(c.ToString()+a);
}
}
}
return comb.ToArray();
}
}

Here's another appraoch that is slighly more generic.
void Main()
{
var perms = new Permutations<char>("abc");
perms.Generate();
}
class Permutations<T> {
private List<T> permutation = new List<T>();
HashSet<T> chosen;
int n;
List<T> sequence;
public Permutations(IEnumerable<T> sequence)
{
this.sequence = sequence.ToList();
this.n = this.sequence.Count;
chosen = new HashSet<T>();
}
public void Generate()
{
if(permutation.Count == n) {
Console.WriteLine(string.Join(",",permutation));
}
else
{
foreach(var elem in sequence)
{
if(chosen.Contains(elem)) continue;
chosen.Add(elem);
permutation.Add(elem);
Generate();
chosen.Remove(elem);
permutation.Remove(elem);
}
}
}
}

How do I 'foreach' through a two-dimensional array?

I've got a two-dimensional array,
string[,] table = {
{ "aa", "aaa" },
{ "bb", "bbb" }
};
And I'd like to foreach through it like this,
foreach (string[] row in table)
{
Console.WriteLine(row[0] + " " + row[1]);
}
But, I get the error:
Can't convert type string to string[]
Is there a way I can achieve what I want, i.e. iterate through the first dimension of the array with the iterator variable returning me the one-dimensional array for that row?

Multidimensional arrays aren't enumerable. Just iterate the good old-fashioned way:
for (int i = 0; i < table.GetLength(0); i++)
{
Console.WriteLine(table[i, 0] + " " + table[i, 1]);
}

As others have suggested, you could use nested for-loops or redeclare your multidimensional array as a jagged one.
However, I think it's worth pointing out that multidimensional arrays are enumerable, just not in the way that you want. For example:
string[,] table = {
{ "aa", "aaa" },
{ "bb", "bbb" }
};
foreach (string s in table)
{
Console.WriteLine(s);
}
/* Output is:
aa
aaa
bb
bbb
*/

If you define your array like this:
string[][] table = new string[][] {
new string[] { "aa", "aaa" },
new string[]{ "bb", "bbb" }
};
Then you can use a foreach loop on it.

UPDATE: I had some time on my hands, so ... I went ahead and fleshed out this idea. See below for the code.
Here's a bit of a crazy answer:
You could do what you're looking for -- essentially treat a two-dimensional array as a table with rows -- by writing a static method (perhaps an extension method) that takes a T[,] and returns an IEnumerable<T[]>. This would require copying each "row" of the underlying table into a new array, though.
A perhaps better (though more involved) approach would be to actually write a class that implements IList<T> as a wrapper around a single "row" of a two-dimensional array (you would probably set IsReadOnly to true and just implement the getter for the this[int] property and probably Count and GetEnumerator; everything else could throw a NotSupportedException). Then your static/extension method could return an IEnumerable<IList<T>> and provide deferred execution.
That way you could write code pretty much like what you have:
foreach (IList<string> row in table.GetRows()) // or something
{
Console.WriteLine(row[0] + " " + row[1]);
}
Just a thought.
Implementation suggestion:
public static class ArrayTableHelper {
public static IEnumerable<IList<T>> GetRows<T>(this T[,] table) {
for (int i = 0; i < table.GetLength(0); ++i)
yield return new ArrayTableRow<T>(table, i);
}
private class ArrayTableRow<T> : IList<T> {
private readonly T[,] _table;
private readonly int _count;
private readonly int _rowIndex;
public ArrayTableRow(T[,] table, int rowIndex) {
if (table == null)
throw new ArgumentNullException("table");
if (rowIndex < 0 || rowIndex >= table.GetLength(0))
throw new ArgumentOutOfRangeException("rowIndex");
_table = table;
_count = _table.GetLength(1);
_rowIndex = rowIndex;
}
// I didn't implement the setter below,
// but you easily COULD (and then set IsReadOnly to false?)
public T this[int index] {
get { return _table[_rowIndex, index]; }
set { throw new NotImplementedException(); }
}
public int Count {
get { return _count; }
}
bool ICollection<T>.IsReadOnly {
get { return true; }
}
public IEnumerator<T> GetEnumerator() {
for (int i = 0; i < _count; ++i)
yield return this[i];
}
// omitted remaining IList<T> members for brevity;
// you actually could implement IndexOf, Contains, etc.
// quite easily, though
}
}
...now I think I should give StackOverflow a break for the rest of the day ;)

It depends on how you define your multi-dimensional array. Here are two options:
using System;
using System.Collections.Generic;
using System.Linq;
namespace ConsoleApplication2
{
class Program
{
static void Main(string[] args)
{
// First
string[,] arr1 = {
{ "aa", "aaa" },
{ "bb", "bbb" }
};
// Second
string[][] arr2 = new[] {
new[] { "aa", "aaa" },
new[] { "bb", "bbb" }
};
// Iterate through first
for (int x = 0; x <= arr1.GetUpperBound(0); x++)
for (int y = 0; y <= arr1.GetUpperBound(1); y++)
Console.Write(arr1[x, y] + "; ");
Console.WriteLine(Environment.NewLine);
// Iterate through second second
foreach (string[] entry in arr2)
foreach (string element in entry)
Console.Write(element + "; ");
Console.WriteLine(Environment.NewLine);
Console.WriteLine("Press any key to finish");
Console.ReadKey();
}
}
}

Here's a simple extension method that returns each row as an IEnumerable<T>. This has the advantage of not using any extra memory:
public static class Array2dExt
{
public static IEnumerable<IEnumerable<T>> Rows<T>(this T[,] array)
{
for (int r = array.GetLowerBound(0); r <= array.GetUpperBound(0); ++r)
yield return row(array, r);
}
static IEnumerable<T> row<T>(T[,] array, int r)
{
for (int c = array.GetLowerBound(1); c <= array.GetUpperBound(1); ++c)
yield return array[r, c];
}
}
Sample usage:
static void Main()
{
string[,] siblings = { { "Mike", "Amy" }, { "Mary", "Albert" }, {"Fred", "Harry"} };
foreach (var row in siblings.Rows())
Console.WriteLine("{" + string.Join(", ", row) + "}");
}

string[][] table = { ... };

string[][] languages = new string[2][];
languages[0] = new string[2];
languages[1] = new string[3];
// inserting data into double dimensional arrays.
for (int i = 0; i < 2; i++)
{
languages[0][i] = "Jagged"+i.ToString();
}
for (int j = 0; j < 3; j++)
{
languages[1][j] = "Jag"+j.ToString();
}
// doing foreach through 2 dimensional arrays.
foreach (string[] s in languages)
{
foreach (string a in s)
{
Console.WriteLine(a);
}
}

Using LINQ you can do it like this:
var table_enum = table
// Convert to IEnumerable<string>
.OfType<string>()
// Create anonymous type where Index1 and Index2
// reflect the indices of the 2-dim. array
.Select((_string, _index) => new {
Index1 = (_index / 2),
Index2 = (_index % 2), // ← I added this only for completeness
Value = _string
})
// Group by Index1, which generates IEnmurable<string> for all Index1 values
.GroupBy(v => v.Index1)
// Convert all Groups of anonymous type to String-Arrays
.Select(group => group.Select(v => v.Value).ToArray());
// Now you can use the foreach-Loop as you planned
foreach(string[] str_arr in table_enum) {
// …
}
This way it is also possible to use the foreach for looping through the columns instead of the rows by using Index2 in the GroupBy instead of Index 1. If you don't know the dimension of your array then you have to use the GetLength() method to determine the dimension and use that value in the quotient.

I'm not a big fan of this method because of the memory usage involved, but if you use the arrays it produces, it isn't such a waste.
public static void ForEachRow<T>(this T[,] list, Action<int, T[]> action)
{
var len = list.GetLength(0);
var sub = list.GetLength(1);
T[] e;
int i, j;
for (i = 0; i < len; i++)
{
e = new T[sub];
for (j = 0; j < sub; j++)
{
e[j] = list[i, j];
}
action(i, e);
}
}
Implementation:
var list = new[,]{0x0, 0x1, 0x2, 0x4, 0x8};
list.ForEachRow((i, row) =>
{
for (var j = 0; j < row.Length; j++)
{
Console.WriteLine("[{0},{1}]: {2}", i, j, row[j]);
}
});
The other solution I found is less memory intensive, but will use more CPU, especially when the dimensions of the arrays' entries are larger.
public static void ForEachRow<T>(this T[,] list, Action<int, IEnumerable<T>> action)
{
var len = list.GetLength(0);
var sub = list.GetLength(1);
int i, j;
IEnumerable<T> e;
for (i = 0; i < len; i++)
{
e = Enumerable.Empty<T>();
for (j = 0; j < sub; j++)
{
e = e.Concat(AsEnumerable(list[i, j]));
}
action(i, e);
}
}
private static IEnumerable<T> AsEnumerable<T>(T add)
{
yield return add;
}
Implementation:
var list = new[,]{0x0, 0x1, 0x2, 0x4, 0x8};
list.ForEachRow((i, row) =>
{
var j = 0;
forrach (var o in row)
{
Console.WriteLine("[{0},{1}]: {2}", i, j, o);
++j;
}
});
As a whole, I find the first option to be more intuitive, especially if you want to access the produced array by its indexer.
At the end of the day, this is all just eye candy, neither methods should really be used in favour of directly accessing the source array;
for (var i = 0; i < list.GetLength(0); i++)
{
foreach (var j = 0; j < list.GetLength(1); j++)
{
Console.WriteLine("[{0},{1}]: {2}", i, j, list[i, j]);
}
}

Remember that a multi-dimensional array is like a table. You don't have an x element and a y element for each entry; you have a string at (for instance) table[1,2].
So, each entry is still only one string (in your example), it's just an entry at a specific x/y value. So, to get both entries at table[1, x], you'd do a nested for loop. Something like the following (not tested, but should be close)
for (int x = 0; x < table.Length; x++)
{
for (int y = 0; y < table.Length; y += 2)
{
Console.WriteLine("{0} {1}", table[x, y], table[x, y + 1]);
}
}

I try this. I hope to help. It work with
static void Main()
{
string[,] matrix = {
{ "aa", "aaa" },
{ "bb", "bbb" }
};
int index = 0;
foreach (string element in matrix)
{
if (index < matrix.GetLength(1))
{
Console.Write(element);
if (index < (matrix.GetLength(1) - 1))
{
Console.Write(" ");
}
index++;
}
if (index == matrix.GetLength(1))
{
Console.Write("\n");
index = 0;
}
}