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Check if a string can be made equal to another string by swapping or replacement of characters

  • Last Updated : 19 May, 2021
Geek Week

Given two strings S1 and S2, the task is to check if string S1 can be made equal to string S2 by swapping any pair of characters replacing any character in the string S1. If it is possible to make the string S1 equal to S2, then print “Yes”. Otherwise, print “No”.

Examples:

Input: S1 = “abc”, S2 = “bca”
Output: Yes
Explanation:
Operation 1: “abc” -> “acb”
Operation 2: “acb” -> “bca”

Input: S1 = “a”, S2 = “aa”
Output: No
Explanation: It is impossible to make the two strings same.

Approach: The idea to solve this problem is to find the frequencies of the characters in the strings and check that the same character is being used in both strings or not. Follow the steps below to solve the problem:



Below is the implementation of this approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find if given strings
// are same or not
bool sameStrings(string str1,
                 string str2)
{
    int N = str1.length();
    int M = str2.length();
 
    // Base Condition
    if (N != M) {
        return false;
    }
 
    // Stores frequency of characters
    // of the string str1 and str2
    int a[256] = { 0 }, b[256] = { 0 };
 
    // Traverse strings str1 & str2 and
    // store frequencies in a[] and b[]
    for (int i = 0; i < N; i++) {
        a[str1[i] - 'a']++;
        b[str2[i] - 'a']++;
    }
 
    // Check if both strings have
    // same characters or not
    int i = 0;
 
    while (i < 256) {
 
        if ((a[i] == 0 && b[i] == 0)
            || (a[i] != 0 && b[i] != 0)) {
            i++;
        }
 
        // If a character is present
        // in one string and is not in
        // another string, return false
        else {
            return false;
        }
    }
 
    // Sort the array a[] and b[]
    sort(a, a + 256);
    sort(b, b + 256);
 
    // Check arrays a and b contain
    // the same frequency or not
    for (int i = 0; i < 256; i++) {
 
        // If the frequencies are not
        // the same after sorting
        if (a[i] != b[i])
            return false;
    }
 
    // At this point, str1 can
    // be converted to str2
    return true;
}
// Driver Code
int main()
{
    string S1 = "cabbba", S2 = "abbccc";
    if (sameStrings(S1, S2))
        cout << "YES" << endl;
    else
        cout << " NO" << endl;
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
class GFG
{
 
// Function to find if given Strings
// are same or not
static boolean sameStrings(String str1,
                 String str2)
{
    int N = str1.length();
    int M = str2.length();
 
    // Base Condition
    if (N != M)
    {
        return false;
    }
 
    // Stores frequency of characters
    // of the String str1 and str2
    int []a = new int[256];
    int []b = new int[256];
 
    // Traverse Strings str1 & str2 and
    // store frequencies in a[] and b[]
    for (int i = 0; i < N; i++)
    {
        a[str1.charAt(i) - 'a']++;
        b[str2.charAt(i) - 'a']++;
    }
 
    // Check if both Strings have
    // same characters or not
    int i = 0;
    while (i < 256)
    {
        if ((a[i] == 0 && b[i] == 0)
            || (a[i] != 0 && b[i] != 0))
        {
            i++;
        }
 
        // If a character is present
        // in one String and is not in
        // another String, return false
        else
        {
            return false;
        }
    }
 
    // Sort the array a[] and b[]
    Arrays.sort(a);
    Arrays.sort(b);
 
    // Check arrays a and b contain
    // the same frequency or not
    for (i = 0; i < 256; i++)
    {
 
        // If the frequencies are not
        // the same after sorting
        if (a[i] != b[i])
            return false;
    }
 
    // At this point, str1 can
    // be converted to str2
    return true;
}
   
// Driver Code
public static void main(String[] args)
{
    String S1 = "cabbba", S2 = "abbccc";
    if (sameStrings(S1, S2))
        System.out.print("YES" +"\n");
    else
        System.out.print(" NO" +"\n");
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 program for the above approach
 
# Function to find if given strings
# are same or not
def sameStrings(str1, str2):
    N = len(str1)
    M = len(str2)
 
    # Base Condition
    if (N != M):
        return False
 
    # Stores frequency of characters
    # of the str1 and str2
    a, b = [0]*256, [0]*256
 
    # Traverse strings str1 & str2 and
    # store frequencies in a[] and b[]
    for i in range(N):
        a[ord(str1[i]) - ord('a')] += 1
        b[ord(str2[i]) - ord('a')] += 1
 
    # Check if both strings have
    # same characters or not
    i = 0
    while (i < 256):
 
        if ((a[i] == 0 and b[i] == 0) or (a[i] != 0 and b[i] != 0)):
            i += 1
 
        # If a character is present
        # in one and is not in
        # another string, return false
        else:
            return False
 
    # Sort the array a[] and b[]
    a = sorted(a)
    b = sorted(b)
 
    # Check arrays a and b contain
    # the same frequency or not
    for i in range(256):
 
        # If the frequencies are not
        # the same after sorting
        if (a[i] != b[i]):
            return False
 
    # At this point, str1 can
    # be converted to str2
    return True
 
  # Driver Code
if __name__ == '__main__':
     
    S1, S2 = "cabbba", "abbccc"
    if (sameStrings(S1, S2)):
        print("YES")
    else:
        print("NO")
 
# This code is contributed by mohit kumar 29.

C#




// C# program for the above approach
using System;
class GFG
{
 
// Function to find if given Strings
// are same or not
static bool sameStrings(string str1,
                 string str2)
{
    int N = str1.Length;
    int M = str2.Length;
 
    // Base Condition
    if (N != M)
    {
        return false;
    }
 
    // Stores frequency of characters
    // of the String str1 and str2
    int []a = new int[256];
    int []b = new int[256];
 
    // Traverse Strings str1 & str2 and
    // store frequencies in a[] and b[]
    for (int j = 0; j < N; j++)
    {
        a[str1[j] - 'a']++;
        b[str2[j] - 'a']++;
    }
 
    // Check if both Strings have
    // same characters or not
    int i = 0 ;
    while (i < 256)
    {
        if ((a[i] == 0 && b[i] == 0)
            || (a[i] != 0 && b[i] != 0))
        {
            i++;
        }
 
        // If a character is present
        // in one String and is not in
        // another String, return false
        else
        {
            return false;
        }
    }
 
    // Sort the array a[] and b[]
    Array.Sort(a);
    Array.Sort(b);
 
    // Check arrays a and b contain
    // the same frequency or not
    for (int j = 0; j < 256; j++)
    {
 
        // If the frequencies are not
        // the same after sorting
        if (a[j] != b[j])
            return false;
    }
 
    // At this point, str1 can
    // be converted to str2
    return true;
}
 
  // Driver Code
  static public void Main()
  {
    string S1 = "cabbba", S2 = "abbccc";
    if (sameStrings(S1, S2))
        Console.Write("YES" +"\n");
    else
        Console.Write(" NO" +"\n");
  }
}
 
// This code is contributed by code_hunt.

Javascript




<script>
      // JavaScript program for the above approach
      // Function to find if given Strings
      // are same or not
      function sameStrings(str1, str2) {
        var N = str1.length;
        var M = str2.length;
 
        // Base Condition
        if (N !== M) {
          return false;
        }
 
        // Stores frequency of characters
        // of the String str1 and str2
        var a = new Array(256).fill(0);
        var b = new Array(256).fill(0);
 
        // Traverse Strings str1 & str2 and
        // store frequencies in a[] and b[]
        for (var j = 0; j < N; j++) {
          a[str1[j].charCodeAt(0) - "a".charCodeAt(0)]++;
          b[str2[j].charCodeAt(0) - "a".charCodeAt(0)]++;
        }
 
        // Check if both Strings have
        // same characters or not
        var i = 0;
        while (i < 256) {
          if ((a[i] === 0 && b[i] === 0) || (a[i] !== 0 && b[i] !== 0)) {
            i++;
          }
 
          // If a character is present
          // in one String and is not in
          // another String, return false
          else {
            return false;
          }
        }
 
        // Sort the array a[] and b[]
        a.sort((x, y) => x - y);
        b.sort((x, y) => x - y);
 
        // Check arrays a and b contain
        // the same frequency or not
        for (var j = 0; j < 256; j++) {
          // If the frequencies are not
          // the same after sorting
          if (a[j] !== b[j]) return false;
        }
 
        // At this point, str1 can
        // be converted to str2
        return true;
      }
 
      // Driver Code
      var S1 = "cabbba",
        S2 = "abbccc";
      if (sameStrings(S1, S2)) document.write("YES" + "<br>");
      else document.write(" NO" + "<br>");
    </script>
Output: 
YES

 

Time Complexity: O(N)
Auxiliary Space: O(256)

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