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Using arrays across multiple classes in C# and insertion sort won't work

I'm working on an assignment for class in which I need to use an insertion sort on an unsorted array, but it won't quite work. The algorithms i sued for the sorts either changed all values to 0 or all to the last value (96) with a 0 on top.
Code for the array class
using System;
namespace UnorderedArrayListNamespace
{
public class UnorderedArrayList
{
public int[] list;
protected int next;
// protected int Count;
public UnorderedArrayList()
{
list = new int[100];
next = 0;
}
public void insert(ref int item)
{
list[next] = item;
next++;
}
public void remove(ref int item)
{
if (next == 0)
{
}
else
{
//find value, if it exists
for (int i = 0; i < next; i++)
{
if (item.Equals(list[i]))
{
for (int j = i; j < next; j++) list[j] = list[j + 1];
next--;
break;
}
}
}
}
public void removeAll(ref int item)
{
}
public int[] InsertionSort(ref int item)
{
for (int i = 0; i < list.Length - 1; i++)
{
for (int j = i + 1; j > 0; j--)
{
if (list[j - 1] > list[j])
{
int temp = list[j - 1];
list[j - 1] = list[j];
list[j] = temp;
}
}
}
return list;
}
public void Minmax(ref int item)
{
int min , max;
min = list[0];
max = list[0];
for (int i = 1; i < list.Length; i++)
{
if (min > list[i])
min = list[i];
if (max < list[i])
max = list[i];
}
Console.WriteLine("min:" + min);
Console.WriteLine("max:" + max);
}
public void print()
{
for (int i = 0; i < next; i++)
{
Console.WriteLine(list[i]);
}
Console.WriteLine();
}
}
}
Code for main class
namespace ConsoleApp3
{
class Program
{
static void Main(string[] args)
{
UnorderedArrayList u = new UnorderedArrayList();
u.print();
int var = 7;
u.insert(ref var);
var = 12;
u.insert(ref var);
var = 5;
u.insert(ref var);
var = 5;
u.insert(ref var);
var = 5;
u.insert(ref var);
var = 96;
u.insert(ref var);
u.print();
var = 5;
u.remove(ref var);
u.print();
u.InsertionSort(ref var);
u.print();
u.Minmax(ref var);
}
}
}
Updated to include all parts of the Array Class. I'm sorry if this is too long winded, it's my first post and I wanted to be thorough.
upon comment request here's the assignment:
Modify the supplied class UnorderedArrayList the following ways:
a. the method remove() removes an element from the list by shifting the elements of the list. However, if the element to be removed is at the beginning of the list and the list is fairly large, it could take a lot of computer time to perform the operation. Because the list elements are in no particular order, you could simply remove the element by copying the last element in the list at the position of the item to be removed and reducing the length of the list.
b. the method remove removes only the first occurrence of an element. Add the method removeAll() that will remove all occurrences of a given element.
c. add the methods min() and max() which will return the smallest and largest respective elements in the list
d. add an insertion sort method that puts the list in order.
e. write a Main() method that thoroughly tests these modifications and demonstrates correctness.

Here you are. in the InsertionSort I first added the item and then did the sort. The minimum is the first item in the list once it is sorted and the maximum the last item. Your MinMax logic was good but just wanted to make you aware once sorted, it is easy to get min,max.
Have a good look at the sorting and how this is done, write down the numbers in the order or the unsorterlist and write down where they change places. This help you understand the logic. Important, you have i & j, both representing the position of items to compare.
7-12- 5- 5-96-5
5-12- 5- 5-96-7 (7 & 5 switched)
5- 5-12- 5-96-7 (12 & 5 switched) etc.
Also, removeAll is same code without the break..
using System;
namespace UnorderedArrayListNamespace
{
public class UnorderedArrayList
{
public int[] list;
protected int next;
// protected int Count;
public UnorderedArrayList()
{
list = new int[100];
next = 0;
}
public void insert(ref int item)
{
list[next] = item;
next++;
}
public void remove(ref int item)
{
//find value, if it exists
for (int i = 0; i < next; i++)
{
if (item.Equals(list[i]))
{
list[i] = list[next - 1]; //Move last item to empty slot
next--;
break; //break because we only remove first
}
}
}
public void removeAll(ref int item)
{
//find value, if it exists
for (int i = 0; i < next; i++)
{
if (item.Equals(list[i]))
{
list[i] = list[next - 1]; //Move last item to empty slot
next--;
}
}
}
public int[] InsertionSort(ref int item)
{
//First we add the item
insert(ref item);
for (int i = 0; i < next - 2; i++)
{
for (int j = next - 1; j > i; j--)
{
if (list[i] > list[j])
{
int temp = list[j];
list[j] = list[i];
list[i] = temp;
}
}
}
return list;
}
public int Min()
{
if (next == 0) throw new Exception("No element is list, cannot get minimum");
return list[0];
}
public int Max()
{
if (next == 0) throw new Exception("No element is list, cannot get maximun");
return list[next - 1];
}
public void print()
{
for (int i = 0; i < next; i++)
{
Console.Write("{0} ", list[i]);
}
Console.WriteLine();
}
}
class Program
{
static void Main(string[] args)
{
UnorderedArrayList u = new UnorderedArrayList();
u.print();
int var = 7;
u.insert(ref var);
var = 12;
u.insert(ref var);
var = 5;
u.insert(ref var);
var = 5;
u.insert(ref var);
var = 5;
u.insert(ref var);
var = 96;
u.insert(ref var);
u.print();
var = 5;
u.remove(ref var);
u.print();
u.InsertionSort(ref var);
u.print();
Console.WriteLine("Minimum of sortedlist is {0}", u.Min());
Console.WriteLine("Maximum of sortedlist is {0}", u.Max());
u.removeAll(ref var);
u.print();
Console.ReadKey();
}
}
}

passing methods working with arrays to main

If I have a method that finds the greatest integer in an array. How do I pass the result back to main?
public static int maxNumber(int[] Array) {
int maxNumber = Array[0];
for (int i = 1; i < Array.length; i++) {
if (List[i] > maxNumber) {
maxNumber = Array[i];
}
return maxNumber;
}
}

Return the result outside the loop, it should be this way
public static int maxNumber(int[] Array)
{
int maxNumber = Array[0];
for (int i = 1; i < Array.length; i++)
{
if (List[i] > maxNumber)
{
maxNumber = Array[i];
}
}
return maxNumber;
}
if you want to get the number as output, call like this
int max = maxNumber(yourArray);

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;
}

Does code exist, for shifting List elements to left or right by specified amount, in C#?

Does code exist, for shifting List elements to left or right by specified amount, in C#?
It is tricky code, it will take some time to write and test special cases, I would rather
reuse something if it exists.
Thanks

Something like this for shift left...
public static void ShiftLeft<T>(List<T> lst, int shifts)
{
for (int i = shifts; i < lst.Count; i++)
{
lst[i - shifts] = lst[i];
}
for (int i = lst.Count - shifts; i < lst.Count; i++)
{
lst[i] = default(T);
}
}
For shift right it's a little more tricky, because we must copy in reverse
public static void ShiftRight<T>(List<T> lst, int shifts)
{
for (int i = lst.Count - shifts - 1; i >= 0; i--)
{
lst[i + shifts] = lst[i];
}
for (int i = 0; i < shifts; i++)
{
lst[i] = default(T);
}
}
With arrays it's a lot more simple, because Array has very powerful methods:
public static void ShiftLeft<T>(T[] arr, int shifts)
{
Array.Copy(arr, shifts, arr, 0, arr.Length - shifts);
Array.Clear(arr, arr.Length - shifts, shifts);
}
public static void ShiftRight<T>(T[] arr, int shifts)
{
Array.Copy(arr, 0, arr, shifts, arr.Length - shifts);
Array.Clear(arr, 0, shifts);
}
And yes, Array.Copy is protected against overleap: If sourceArray and destinationArray overlap, this method behaves as if the original values of sourceArray were preserved in a temporary location before destinationArray is overwritten.

Below are a couple of extension methods that will shift the list either right or left. The methods will return a list.
public static class ShiftList
{
public static List<T> ShiftLeft<T>(this List<T> list, int shiftBy)
{
if (list.Count <= shiftBy)
{
return list;
}
var result = list.GetRange(shiftBy, list.Count-shiftBy);
result.AddRange(list.GetRange(0,shiftBy));
return result;
}
public static List<T> ShiftRight<T>(this List<T> list, int shiftBy)
{
if (list.Count <= shiftBy)
{
return list;
}
var result = list.GetRange(list.Count - shiftBy, shiftBy);
result.AddRange(list.GetRange(0, list.Count - shiftBy));
return result;
}
}
Here's an example of how to call it.
class Program
{
static void Main(string[] args)
{
List<int> test = Enumerable.Range(0, 10).ToList();
test = test.ShiftLeft(1);
PrintList(test);
Console.WriteLine("");
PrintList(test.ShiftRight(2));
Console.ReadLine();
}
private static void PrintList(List<int> test)
{
for (int i = 0; i < test.Count; i++)
{
Console.WriteLine(test[i]);
}
}
}

Keep it simple by taking the first part and second part and flipping them. Same thing but flip other way for the ShiftRight
public static List<int> ShiftLeft(List<int> a, int d)
{
if (a.Count > d)
{
var beginingPart = a.GetRange(0, d);
var remainingPart = a.GetRange(d, a.Count - d);
return remainingPart.Concat(beginingPart).ToList();
}
else if (a.Count < d)
{
var mod = d % a.Count;
if (mod != 0)
{
return rotLeft(a, mod);
}
}
return a;
}

Given that "iterations" is the times you want to shift and "numbers" is the List
Shift left:
static void ShiftLeft(int iterations)
{
for (int i = 0; i < iterations; i++)
{
numbers.Add(numbers[0]);
numbers.RemoveAt(0);
}
}
ShiftRight:
static void ShiftRight(int iterations)
{
for (int i = 0; i < iterations; i++)
{
numbers.Insert(0, numbers[numbers.Count - 1]);
numbers.RemoveAt(numbers.Count - 1);
}
}