Checking if two strings are permutations of each other

Question

How to determine if two strings are permutations of each other

Answer 1

    public static boolean isPermutation(String s1, String s2) {
    if (s1.length() != s2.length()) {
        return false;
    }else if(s1.length()==0 ){
        return true;
    }
    else if(s1.length()==1){
        return s1.equals(s2);
    }
    char[] s = s1.toCharArray();
    char[] t = s2.toCharArray();

    for (int i = 0; i < s.length; i++) {
        for (int j = 0; j < t.length; j++) {

            if (s.length == s1.length() && (i == 0 && j == t.length - 1) && s[i] != t[j]) {
                return false;
            }
            if (s[i] == t[j]) {
                String ss = new String(s);
                String tt = new String(t);
                s = (ss.substring(0, i) + ss.substring(i + 1, s.length)).toCharArray();
                t = (tt.substring(0, j) + tt.substring(j + 1, t.length)).toCharArray();

                System.out.println(new String(s));
                System.out.println(new String(t));

                i = 0;
                j = 0;
            }
        }
    }
    return s[0]==t[0] ;
}

This solution works for any charset. With an O(n) complexity.

The output for: isPermutation("The Big Bang Theory", "B B T Tehgiangyroeh")

he Big Bang Theory
B B  Tehgiangyroeh
e Big Bang Theory
B B  Tegiangyroeh
 Big Bang Theory
B B  Tgiangyroeh
Big Bang Theory
BB  Tgiangyroeh
ig Bang Theory
B  Tgiangyroeh
g Bang Theory
B  Tgangyroeh
 Bang Theory
B  Tangyroeh
Bang Theory
B Tangyroeh
Bng Theory
B Tngyroeh
Bg Theory
B Tgyroeh
B Theory
B Tyroeh
BTheory
BTyroeh
Bheory
Byroeh
Beory
Byroe
Bory
Byro
Bry
Byr
By
By
B
B
true

Answer 2

To put @Michael Borgwardt's words in to code:

public boolean checkAnagram(String str1, String str2) {

    if (str1.length() != str2.length())
      return false;

    char[] a = str1.toCharArray();
    char[] b = str2.toCharArray();

    Arrays.sort(a);
    Arrays.sort(b);

    return Arrays.equals(a, b);
}

Answer 3

public boolean permitation(String s,String t){
      if(s.length() != t.length()){
          return false;
          }

       int[] letters = new int[256];//Assumes that the character set is ASCII
       char[] s_array = s.toCharArray();

       for(char c:s_array){         /////count number of each char in s
             letters[c]++;
        }
       for(int i=0;i<t.length();i++){
             int c = (int)t.charAt(i);
             if(--letters[c]<0){
                  return false;
             }
        }
        return true;
}

Answer 4

>>>def isPermutation = lambda x, y: set([i for i in x]) == set([j for j in x])
>>>isPermutation("rasp", "spar")
>>>True

Answer 5

It can be done by using a Dictionary in C#. A basic implementation is like :

    private static bool IsPermutation(string str1, string str2)
    {
        if (str1.Length != str2.Length)
            return false;

        var dictionary = new Dictionary<char, int>();

        foreach(var x in str1)
        {
            if (dictionary.ContainsKey(x))
                dictionary[x] += 1;
            else
                dictionary.Add(x, 1);
        }

        foreach (var y in str2)
        {
            if (dictionary.ContainsKey(y))
            {
                if (dictionary[y] > 0)
                    dictionary[y] -= 1;
                else
                    return false;
            }
            else
                return false;
        }

        foreach(var el in dictionary)
        {
            if (el.Value != 0)
                return false;
        }

        return true;
    }

Time Complexity is O(n), linear solution.

Answer 6

I wanted to give a recursive solution for this problem as I did not find any answer which was recursive. I think that this code seems to be tough but if you'll try to understand it you'll see the beauty of this code. Recursion is sometimes hard to understand but good to use! This method returns all the permutations of the entered String 's' and keeps storing them in the array 'arr[]'. The value of 't' initially is blank "" .

import java.io.*;
class permute_compare2str
{
    static String[] arr= new String [1200];
    static int p=0;
    void permutation(String s,String t)
    {
        if (s.length()==0)
        {
            arr[p++]=t;
            return;
        }
        for(int i=0;i<s.length();i++)
            permutation(s.substring(0,i)+s.substring(i+1),t+s.charAt(i));
    }
    public static void main(String kr[])throws IOException
    {
        int flag = 0;
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        System.out.println("Enter the first String:");
        String str1 = br.readLine();
        System.out.println("Enter the second String:");
        String str2 = br.readLine();
        new permute_compare2str().permutation(str1,"");
        for(int i = 0; i < p; ++i)
        {
            if(arr[i].equals(str2))
            {
                flag = 1;
                break;
            }
        }
        if(flag == 1)
            System.out.println("True");
        else
        {
            System.out.println("False");
            return;
        }
    }
}

One limitation that I can see is that the length of the array is fixed and so will not be able to return values for a large String value 's'. Please alter the same as per the requirements. There are other solutions to this problem as well.

I have shared this code because you can actually use this to get the permutations of a string printed directly without the array as well.

HERE:

void permutations(String s, String t)
    {
        if (s.length()==0)
        {
            System.out.print(t+" ");
            return;
        }
        for(int i=0;i<s.length();i++)
            permutations(s.substring(0,i)+s.substring(i+1),t+s.charAt(i));
    }

Value of 's' is the string whose permutations is needed and value of 't' is again empty "".

Reference: Introduction to Programming in Java