Develop Reference

2D array 3x9 matrix get the index of the next two elements of the current index and check if it is equal to 0

I am a total newbie to programming and i have been following some tutorials on array related to housie ticket generator.The point where I am stuck is that, I have to check each rows and each rows of the 3x9 matrix should not have more the two empty cells or it cannot have more then two cells filled next to each other.I am putting random numbers on the arrays and trying to validate the rules but,the program crashes. Can someone please give me an idea.?
This is what i've tried.
for(int columnIndex=0;columnIndex<=6;columnIndex++)
{
if(game[i,columnIndex+2]!=0)
{
return -1;
}
}
And this is the whole code
using System;
namespace HelloWorld
{
class Program
{
public static void Main (String[] args)
{
for(int times=0;times<2;times++)
{
startGame();
Console.WriteLine("******************************************************************");
}
}
private static void startGame()
{
int[,] game = new int[3, 9];
int occupancyLimit = 15;
while (occupancyLimit > 0)
{
int i = getRandomNumber(3);
int j = getRandomNumber(9);
//Console.Write(i);
//Console.Write(j);
// Console.Write(game[i,j]+" ");
int data = validateAndReturnNumber(i, j, game);
if (data>0)
{
game[i, j] = data;
occupancyLimit--;
//Console.WriteLine(game[i,j]);
}
}
for (int i = 0; i < game.GetLength(0); i++)
{
for (int j = 0; j < game.GetLength(1); j++)
{
Console.Write(game[i,j] + "\t");
}
Console.WriteLine();
}
}
private static int validateAndReturnNumber(int i, int j, int[,] game)
{
//do not override existing elements in array
if (game[i,j] != 0)
{
return -1;
}
//column cannot have more then two elements
int columncounter = 0;
for(int c=0;c<3;c++)
{
if(game[c,j]!=0)
{
columncounter++;
}
}
if(columncounter>=2)
{
return -1;
}
/*
//rows cannot have more then two cells filled in and
//rows cannot have more then two empty cells
for(int columnIndex=0;columnIndex<=6;columnIndex++)
{
if(game[i,columnIndex+2]!=0)
{
return -1;
}
}
*/
// rows cannot have more then 5 elements
int rowcounter = 0;
for(int r=0;r<9;r++)
{
if(game[i,r]!=0)
{
rowcounter++;
}
}
//Applying, rows cannot have more then 5 elements
if(rowcounter>=5)
{
return -1;
}
//return getRandomNumberForColumn(j);
int data = 0;
Boolean isValueSet = false;
do
{
data = getRandomNumberForColumn(j);
isValueSet = isValueExistsInCol(game, i, j, data);
} while (isValueSet);
return data;
}
private static bool isValueExistsInCol(int[,] game, int i, int j, int data)
{
Boolean status = false;
for(int k=0;k<3;k++)
{
if(game[k,j]==data)
{
status = true;
break;
}
}
return status;
}
private static int getRandomNumberForColumn(int high)
{
if(high==0)
{
high = 10;
}
else
{
high=(high + 1) * 10;
}
int low = high - 9;
Random random = new Random();
return random.Next(high-low)+low;
}
private static int getRandomNumber(int max)
{
Random random = new Random();
int num=random.Next(max);
return (num);
}
}
}

Why your for loop does not work:
for (int columnIndex = 0; columnIndex <= 6; columnIndex++)
{
if (game[i, columnIndex + 2] != 0)
{
return -1;
}
}
This loop does not take j into account. It is testing for previous numbers added, as soon as one previous number fails this test, it will fail indefinitely. This creates an infinite loop. This loop also fails if a number is placed in any position past 1, while it needs to fill 5 positions to succeed. This is mathematically impossible.
This:
'should not have more the two empty cells or it cannot have more then two cells' is also mathematically impossible. a row of 9 cannot have less than 2 full and less than 2 empty at the same time.
I think you are after 2 empty or full consecutively in a row. For that testing for two empty in a row cannot be achieved as it starts empty, and you are testing it before you fill it. Fortunately this is a redundant test that will always be true based on all of the other test cases.
No more than 2 full in a row is possible, but can also lead to impossible scenarios. I have added a check that resets the scenario if it has not found the solution after 1000 guesses.
using System;
namespace HelloWorld
{
class Program
{
public static void Main(String[] args)
{
for (int times = 0; times < 2; times++)
{
startGame();
// Console.WriteLine("******************************************************************");
}
}
private static void startGame()
{
int iCount = 0;
int[,] game = new int[3, 9];
int occupancyLimit = 15;
while (occupancyLimit > 0)
{
int i = getRandomNumber(3);
int j = getRandomNumber(9);
//Console.Write(i);
//Console.Write(j);
// Console.Write(game[i,j]+" ");
int data = validateAndReturnNumber(i, j, game);
if (data > 0)
{
game[i, j] = data;
occupancyLimit--;
//Console.WriteLine(game[i,j]);
}
else
{
iCount++;
//Console.WriteLine(iCount);
//printGame(game);
// If X many fails, retry
if(iCount > 1000)
{
iCount = 0;
game = new int[3, 9];
occupancyLimit = 15;
}
}
// If you want to check for zeros you would need to do it here. And use while(true) above
/*
if( //Check for zeros)
{
break; // Ends While loop
}
else
{
// Reset and try again
iCount = 0;
game = new int[3, 9];
occupancyLimit = 15;
}
*/
}
printGame(game);
}
private static void printGame(int[,] game)
{
for (int i = 0; i < game.GetLength(0); i++)
{
for (int j = 0; j < game.GetLength(1); j++)
{
Console.Write(game[i, j] + "\t");
}
Console.WriteLine();
}
Console.WriteLine("******************************************************************");
}
private static int validateAndReturnNumber(int i, int j, int[,] game)
{
//do not override existing elements in array
if (game[i, j] != 0)
{
return -1;
}
//column cannot have more then two elements
int columncounter = 0;
for (int c = 0; c < 3; c++)
{
if (game[c, j] != 0)
{
columncounter++;
}
}
if (columncounter >= 2)
{
return -1;
}
if(
(j != 0 && j != 1 && game[i, j - 2] != 0 && game[i, j - 1] != 0) || // 12X
(j != 0 && j != 8 && game[i, j - 1] != 0 && game[i, j + 1] != 0) || // 1X3
(j != 7 && j != 8 && game[i, j + 1] != 0 && game[i, j + 2] != 0) // X23
)
{
return -1;
}
//for (int columnIndex = 0; columnIndex <= 6; columnIndex++)
//{
// if (game[i, columnIndex + 2] != 0)
// {
// return -1;
// }
//}
// rows cannot have more then 5 elements
int rowcounter = 0;
for (int r = 0; r < 9; r++)
{
if (game[i, r] != 0)
{
rowcounter++;
}
}
//Applying, rows cannot have more then 5 elements
if (rowcounter >= 5)
{
return -1;
}
//return getRandomNumberForColumn(j);
int data = 0;
Boolean isValueSet = false;
do
{
data = getRandomNumberForColumn(j);
isValueSet = isValueExistsInCol(game, i, j, data);
} while (isValueSet);
return data;
}
private static bool isValueExistsInCol(int[,] game, int i, int j, int data)
{
Boolean status = false;
for (int k = 0; k < 3; k++)
{
if (game[k, j] == data)
{
status = true;
break;
}
}
return status;
}
private static int getRandomNumberForColumn(int high)
{
if (high == 0)
{
high = 10;
}
else
{
high = (high + 1) * 10;
}
int low = high - 9;
Random random = new Random();
return random.Next(high - low) + low;
}
private static int getRandomNumber(int max)
{
Random random = new Random();
int num = random.Next(max);
return (num);
}
}
}
Cheers!

C# Anagram Checker With LinkedList

I'm trying to check if two words are anagram and trying to do this with LinkedList.To do that,first,I created a class named LinkedList:
class LinkedList
{
private Node head;
private int count;
public LinkedList()
{
this.head = null;
this.count = 0;
}
public bool Empty
{
get { return this.count == 0; }
}
public int Count
{
get { return this.count; }
}
public object this[int index]
{
get { return this.Get(index); }
}
public object Add(int index,object o)
{
if (index < 0)
{
throw new ArgumentOutOfRangeException("Index - " + index); //if index is less than 0 throw an error message
}
if (index > count) // if size exceeds the limit of the list the item will be added to the last line of the list.
{
index = count;
}
Node current = this.head;
if(this.Empty || index== 0)
{
this.head = new Node(o, this.head);
}
else
{
for(int i = 0; i < index - 1; i++)
{
current = current.Next;
}
current.Next = new Node(o, current.Next);
}
count++;
return o;
}
public object Add(Object o)
{
return this.Add(count, o);
}
public object Remove(int index)
{
if (index < 0)
{
throw new ArgumentOutOfRangeException("Index - " + index);
}
if (this.Empty)
{
return null;
}
if (index >= this.count)
{
index = count-1;
}
Node current = this.head;
object result = null;
if (index == 0)
{
result = current.Data; //gets the first node
this.head = current.Next; //makes 2nd node to the first node
}
else
{
for(int i = 0; i < index - 1; i++)
{
result = current.Next.Data;
}
result = current.Next;
current.Next = current.Next.Next;
}
count--;
return result;
}
public int IndexOf(object o)
{
Node current = this.head;
for(int i = 0; i < this.count; i++)
{
if (current.Data.Equals(o))
{
return i;
}
current = current.Next;
}
return -1;
}
public bool Contains(object o)
{
return this.IndexOf(o) >= 0; //if list contains object it returns bigger value than -1 and also 0.
}
public object Get(int index)
{
if(index < 0)
{
throw new ArgumentOutOfRangeException("Index - " + index);
}
if (this.Empty)
{
return null;
}
if(index >= this.count)
{
index = this.count-1;
}
Node current = this.head;
for(int i=0;i< index; i++)
{
current = current.Next;
}
return current.Data;
}
}
And another class named "Node":
class Node
{
private object data;
private Node next;
public Node(object data,Node next) //constructor
{
this.data = data;
this.next = next;
}
public object Data
{
get { return this.data; }
set { this.data = value; }
}
public Node Next
{
get { return this.next; }
set { this.next = value; }
}
}
And in main program,I created two objects from linkedlist class and read two strings from user and added the words' chars to the linked list.And compared the chars and if they found they'll be deleted from linkedlist,increases the counter and so on.If counter equals to the list1's number ofelements then they are anagrams if not the words are not anagrams.Here's my main program code:
class Program
{
static void Main(string[] args)
{
int counter = 0;
String word1, word2;
Console.WriteLine("Welcome to Anagram Checker!\nPlease enter your first word:");
word1 = Console.ReadLine();
Console.WriteLine("\nPlease enter the second word:");
word2 = Console.ReadLine();
int result = AnagramChecker(word1, word2, counter);
if (result == 1)
{
Console.WriteLine("These words are anagram");
}
if (result == 0)
{
Console.WriteLine("The words are not anagrams");
}
Console.ReadLine();
}
public static int AnagramChecker(String word1, String word2, int counter)
{
char[] ArrayWord1 = word1.ToCharArray();
char[] ArrayWord2 = word2.ToCharArray();
LinkedList list1 = new LinkedList();
LinkedList list2 = new LinkedList();
for (int i = 0; i < ArrayWord1.Length; i++) //Adds char of word1 to the list
{
list1.Add(i,ArrayWord1[i]);
}
for (int j = 0; j < ArrayWord2.Length; j++) //Adds char of word2 to the list
{
list2.Add(j,ArrayWord2[j]);
}
int max;
if (list1.Count >= list2.Count)
{
max = list1.Count;
}
if (list2.Count > list1.Count)
{
max = list2.Count;
}
for (int i = 0; i < list1.Count; i++)
{
if (list2.Contains(list1[i]) && list1.Contains(list2[i]))
{
list1.Remove(i);
list2.Remove(list2.IndexOf(list1[i]));
counter++;
}
}
Console.WriteLine(counter);
if (counter == word1.Length || counter == word2.Length)
{
return 1;
}
else
return 0;
}
}
When I'm entering different words I get different results.The output examples are below.What do I do wrong?
Outputs:
1-)
2-)
Thanks for your kind helps.

If you're just looking to find if words are anagrams, you can use this method:
private static bool areAnagrams(string word1, string word2)
{
List<char> w1 = word1.OrderBy(c => c).ToList();
List<char> w2 = word2.OrderBy(c => c).ToList();
return !w1.Where((t, i) => t != w2[i]).Any();
}
Which create two lists ordered with the words chars, then compare both.
More readable equivalent:
private static bool areAnagrams(string word1, string word2)
{
List<char> w1 = word1.OrderBy(c => c).ToList();
List<char> w2 = word2.OrderBy(c => c).ToList();
if (w1.Count != w2.Count)
return false;
for (int i = 0; i < w1.Count; i++)
{
if (w1[i] != w2[i])
return false;
}
return true;
}

I fixed the problem,I just modified the if statement which checks if they're anagram or not:
for (int i = 0; i < list1.Count; i++)
{
if (list2.Contains(list1[i]))
{
list1.Remove(i);
// list2.Remove(list2.IndexOf(list1[i]));
i--;
counter++;
}
Thanks all of you for your helps :)

Your answers are:
int[] sayilar1 = new int[150];
int[] sayilar2 = new int[150];
public Form1()
{
InitializeComponent();
}
private void Form1_Load(object sender, EventArgs e)
{
var rand = new Random();
for (int i = 0; i < sayilar1.Length; i++)
{
sayilar1[i] = rand.Next();
sayilar2[i] = rand.Next();
lvNumbers.Items.Add(sayilar1[i].ToString());
lvNumbers.Items.Add(sayilar2[i].ToString());
}
}
private void btnShuffle_Click(object sender, EventArgs e)
{
int[]newArray=BubbleSort();
for (int i = 0; i < newArray.Count(); i++)
{
lvSorted.Items.Add(newArray[i].ToString());
}
}
private int[] BubbleSort()
{
int temp = 0;
int[] newArray = new int[300];
for (int i = 0; i < 300; i++)
{
if (i < 150)
{
newArray[i] = sayilar1[i];
}
if (i >= 150)
newArray[i] = sayilar2[i - 150];
}
for (int i = 0; i < newArray.Length; i++)
{
for (int sort = 0; sort < newArray.Length - 1; sort++)
{
if (newArray[sort] > newArray[sort + 1])
{
temp = newArray[sort + 1];
newArray[sort + 1] = newArray[sort];
newArray[sort] = temp;
}
}
}
return newArray;
}
}
2.
private void btnTek_Click(object sender, EventArgs e)
{
lvFiltered.Items.Clear();
string[] sayilar = tbSayilar.Text.Split('\n');
int[] array = new int[sayilar.Length];
for (int i = 0; i < sayilar.Length; i++)
{
array[i] = int.Parse(sayilar[i]);
}
List<int> ayiklanmisSayilar = TekCiftAyir(array, "T");
for (int i = 0; i < ayiklanmisSayilar.Count; i++)
{
lvFiltered.Items.Add(ayiklanmisSayilar[i].ToString());
}
}
private void btnCift_Click(object sender, EventArgs e)
{
lvFiltered.Items.Clear();
string[] sayilar = tbSayilar.Text.Split('\n');
int[] array = new int[sayilar.Length];
for (int i = 0; i < sayilar.Length; i++)
{
array[i] = int.Parse(sayilar[i]);
}
List<int> ayiklanmisSayilar = TekCiftAyir(array, "C");
for (int i = 0; i < ayiklanmisSayilar.Count; i++)
{
lvFiltered.Items.Add(ayiklanmisSayilar[i].ToString());
}
}
private List<int> TekCiftAyir(int[] array, string TC)
{
List<int> ayiklanmisSayilar = new List<int>();
if (TC == "T")
{
for (int i = 0; i < array.Length; i++)
{
if (array[i] % 2 == 1)
{
ayiklanmisSayilar.Add(array[i]);
}
}
}
if (TC == "C")
{
for (int i = 0; i < array.Length; i++)
{
if (array[i] % 2 == 0)
{
ayiklanmisSayilar.Add(array[i]);
}
}
}
return ayiklanmisSayilar;
}
}

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;
}

converting from breadth first search to depth first limited search

as all newbies to the searching algorithms, i am working on an example of the old fashion 8-problem puzzle, i have done the breadth first algorithm which is in the code below, and i was wondering how can convert it to the depth first limited search.
how can I convert from the breadth first algorithm to the depth first limited search algorithm??
code :
class BFS
{
int[] GoalState = { 0, 1, 2, 3, 4, 5, 6, 7, 8 };
private Queue<Node> Frontier;
private HashSet<Node> Solution;
private Node RootNode;
public BFS(int[] StartState)
{
this.RootNode = new Node(StartState, -1, "\0");
Solution = new HashSet<Node>();
Frontier = new Queue<Node>();
}
public void Solve()
{
Node ActiveNode = RootNode;
bool IsSolved = false;
while (!IsGoalState(ActiveNode.GetState()))
{
Solution.Add(ActiveNode);
foreach (Node successor in GenerateSuccessor(ActiveNode))
{
Frontier.Enqueue(successor);
if (IsGoalState(successor.GetState()))
{
IsSolved = true;
Solution.Add(successor);
break;
}
}
if (IsSolved)
break;
ActiveNode = Frontier.Dequeue();
}
WriteSolution();
}
public IEnumerable<Node> GenerateSuccessor(Node ParentNode)
{
int[] ParentState = ParentNode.GetState();
int EmptySpacePosition = 0;
int temp;
for (int x = 0; x < 9; x++)
{
if (ParentState[x] == 0)
{
EmptySpacePosition = x;
break;
}
}
//Can move empty space to LEFT?
if (EmptySpacePosition != 0 && EmptySpacePosition != 3 && EmptySpacePosition != 6)
{
int[] _State = (int[])ParentState.Clone();
_State[EmptySpacePosition] = _State[EmptySpacePosition - 1];
_State[EmptySpacePosition - 1] = 0;
yield return new Node(_State, ParentNode.GetId(), "Left ");
}
//Can move empty space to RIGHT?
if (EmptySpacePosition != 2 && EmptySpacePosition != 5 && EmptySpacePosition != 8)
{
int[] _State = (int[])ParentState.Clone();
_State[EmptySpacePosition] = _State[EmptySpacePosition + 1];
_State[EmptySpacePosition + 1] = 0;
yield return new Node(_State, ParentNode.GetId(),"Right ");
}
//Can move empty space to UP?
if (EmptySpacePosition > 2)
{
int[] _State = (int[])ParentState.Clone();
_State[EmptySpacePosition] = _State[EmptySpacePosition - 3];
_State[EmptySpacePosition - 3] = 0;
yield return new Node(_State, ParentNode.GetId(), "Up ");
}
//Can move empty space to DOWN?
if (EmptySpacePosition < 6)
{
int[] _State = (int[])ParentState.Clone();
_State[EmptySpacePosition] = _State[EmptySpacePosition + 3];
_State[EmptySpacePosition + 3] = 0;
yield return new Node(_State, ParentNode.GetId(),"Down ");
}
}
public bool IsGoalState(int[] State)
{
for (int x = 0; x < 9; x++)
{
if (State[x] != GoalState[x])
return false;
}
return true;
}
public void WriteSolution()
{
StringBuilder s = new StringBuilder();
int ParentId = 0;
#region InfoPrint
Console.Write("Puzzle= ");
foreach (int i in RootNode.GetState())
{
Console.Write(i);
}
Console.WriteLine();
Console.WriteLine("Nodes Generated= " + (Solution.Count + Frontier.Count));
#endregion
foreach (Node n in Solution.Reverse())
{
if (ParentId == 0 || n.GetId() == ParentId)
{
s.Append(n.GetMove());
ParentId = n.GetParentId();
}
}
Console.WriteLine("Solution Length= " + (s.Length - 1));
Console.WriteLine("Solution= " + s.ToString());
}
}
and here is my class node
class Node
{
static int _IdCnt = 0;
private int[] State;
private int Id;
private int ParentId;
private string Move;
public Node(int[] State, int ParentId, string Move)
{
Id = _IdCnt++;
this.State = State;
this.ParentId = ParentId;
this.Move = Move;
}
public void SetState(int[] State)
{
this.State = State;
}
public int[] GetState()
{
return (int[])State.Clone();
}
public int GetId()
{
return Id;
}
public int GetParentId()
{
return ParentId;
}
public string GetMove()
{
return Move;
}
}

In the DFS, you'll need to use stack instead of queue. Basically, you add the root node to a stack, and while stack contains a node, you pop the node, do your logic, add its neighbors to the stack and continue.
1. Add root to a stack.
2. while(stack.Count > 0)
{
3. pop the stack
4. if matches, return
5. else add neighbors to stack
}
return not found