Data Modelling & AI Data Structure & Algorithm

Map every character of one string to another such that all occurrences are mapped to the same character

21 June 2025

0

Given two strings s1 and s2, the task is to check if characters of the first string can be mapped with the character of the second string such that if a character ch1 is mapped with some character ch2 then all the occurrences of ch1 will only be mapped with ch2 for both the strings.

Examples:

Input: s1 = "axx", s2 = "cbc" 
Output: Yes 
'a' in s1 can be mapped to 'b' in s2 
and 'x' in s1 can be mapped to 'c' in s2.

Input: s1 = "a", s2 = "df" 
Output: No

Approach: If the lengths of both the strings are unequal then the strings cannot be mapped else create two frequency arrays freq1[] and freq2[] which will store the frequencies of all the characters of the given strings s1 and s2 respectively. Now, for every non-zero value in freq1[] find an equal value in freq2[]. If all the non-zero values from freq1[] can be mapped to some value in freq2[] then the answer is possible else not.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
#define MAX 26
 
// Function that returns true if the mapping is possible
bool canBeMapped(string s1, int l1, string s2, int l2)
{
 
    // Both the strings are of un-equal lengths
    if (l1 != l2)
        return false;
 
    // To store the frequencies of the
    // characters in both the string
    int freq1[MAX] = { 0 };
    int freq2[MAX] = { 0 };
 
    // Update frequencies of the characters
    for (int i = 0; i < l1; i++)
        freq1[s1[i] - 'a']++;
    for (int i = 0; i < l2; i++)
        freq2[s2[i] - 'a']++;
 
    // For every character of s1
    for (int i = 0; i < MAX; i++) {
 
        // If current character is
        // not present in s1
        if (freq1[i] == 0)
            continue;
        bool found = false;
 
        // Find a character in s2 that has frequency
        // equal to the current character's
        // frequency in s1
        for (int j = 0; j < MAX; j++) {
 
            // If such character is found
            if (freq1[i] == freq2[j]) {
 
                // Set the frequency to -1 so that
                // it doesn't get picked again
                freq2[j] = -1;
 
                // Set found to true
                found = true;
                break;
            }
        }
 
        // If there is no character in s2
        // that could be mapped to the
        // current character in s1
        if (!found)
            return false;
    }
 
    return true;
}
 
// Driver code
int main()
{
    string s1 = "axx";
    string s2 = "cbc";
    int l1 = s1.length();
    int l2 = s2.length();
 
    if (canBeMapped(s1, l1, s2, l2))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java

// Java implementation of the approach 
import java.io.*;
public class GFG
{
 
    static int MAX = 26;
 
    // Function that returns true if the mapping is possible
    public static boolean canBeMapped(String s1, int l1, 
                                        String s2, int l2) 
    {
         
        // Both the strings are of un-equal lengths
        if (l1 != l2)
            return false;
 
        // To store the frequencies of the
        // characters in both the string
        int[] freq1 = new int[MAX];
        int[] freq2 = new int[MAX];
 
        // Update frequencies of the characters
        for (int i = 0; i < l1; i++)
            freq1[s1.charAt(i) - 'a']++;
        for (int i = 0; i < l2; i++)
            freq2[s2.charAt(i) - 'a']++;
 
        // For every character of s1
        for (int i = 0; i < MAX; i++) {
 
            // If current character is
            // not present in s1
            if (freq1[i] == 0)
                continue;
            boolean found = false;
 
            // Find a character in s2 that has frequency
            // equal to the current character's
            // frequency in s1
            for (int j = 0; j < MAX; j++) 
            {
 
                // If such character is found
                if (freq1[i] == freq2[j]) 
                {
 
                    // Set the frequency to -1 so that
                    // it doesn't get picked again
                    freq2[j] = -1;
 
                    // Set found to true
                    found = true;
                    break;
                }
            }
 
            // If there is no character in s2
            // that could be mapped to the
            // current character in s1
            if (!found)
                return false;
        }
 
        return true;
    }
 
    // Driver code
    public static void main(String[] args) 
    {
        String s1 = "axx";
        String s2 = "cbc";
        int l1 = s1.length();
        int l2 = s2.length();
 
        if (canBeMapped(s1, l1, s2, l2))
            System.out.println("Yes");
        else
            System.out.println("No");
 
    }
}
 
// This code is contributed by
// sanjeev2552

Python3

# Python 3 implementation of the approach
 
MAX = 26
 
# Function that returns true if the mapping is possible
def canBeMapped(s1, l1, s2, l2):
    # Both the strings are of un-equal lengths
    if (l1 != l2):
        return False
 
    # To store the frequencies of the
    # characters in both the string
    freq1 = [0 for i in range(MAX)]
    freq2 = [0 for i in range(MAX)]
 
    # Update frequencies of the characters
    for i in range(l1):
        freq1[ord(s1[i]) - ord('a')] += 1
    for i in range(l2):
        freq2[ord(s2[i]) - ord('a')] += 1
 
    # For every character of s1
    for i in range(MAX):
        # If current character is
        # not present in s1
        if (freq1[i] == 0):
            continue
        found = False
 
        # Find a character in s2 that has frequency
        # equal to the current character's
        # frequency in s1
        for j in range(MAX):
            # If such character is found
            if (freq1[i] == freq2[j]):
                # Set the frequency to -1 so that
                # it doesn't get picked again
                freq2[j] = -1
 
                # Set found to true
                found = True
                break
 
        # If there is no character in s2
        # that could be mapped to the
        # current character in s1
        if (found==False):
            return False
 
    return True
 
# Driver code
if __name__ == '__main__':
    s1 = "axx"
    s2 = "cbc"
    l1 = len(s1)
    l2 = len(s2)
 
    if (canBeMapped(s1, l1, s2, l2)):
        print("Yes")
    else:
        print("No")
         
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach 
using System;
     
class GFG
{
    static int MAX = 26;
 
    // Function that returns true
    // if the mapping is possible
    public static Boolean canBeMapped(String s1, int l1, 
                                      String s2, int l2) 
    {
         
        // Both the strings are of un-equal lengths
        if (l1 != l2)
            return false;
 
        // To store the frequencies of the
        // characters in both the string
        int[] freq1 = new int[MAX];
        int[] freq2 = new int[MAX];
 
        // Update frequencies of the characters
        for (int i = 0; i < l1; i++)
            freq1[s1[i] - 'a']++;
        for (int i = 0; i < l2; i++)
            freq2[s2[i] - 'a']++;
 
        // For every character of s1
        for (int i = 0; i < MAX; i++)
        {
 
            // If current character is
            // not present in s1
            if (freq1[i] == 0)
                continue;
            Boolean found = false;
 
            // Find a character in s2 that has frequency
            // equal to the current character's
            // frequency in s1
            for (int j = 0; j < MAX; j++) 
            {
 
                // If such character is found
                if (freq1[i] == freq2[j]) 
                {
 
                    // Set the frequency to -1 so that
                    // it doesn't get picked again
                    freq2[j] = -1;
 
                    // Set found to true
                    found = true;
                    break;
                }
            }
 
            // If there is no character in s2
            // that could be mapped to the
            // current character in s1
            if (!found)
                return false;
        }
        return true;
    }
 
    // Driver code
    public static void Main(String[] args) 
    {
        String s1 = "axx";
        String s2 = "cbc";
        int l1 = s1.Length;
        int l2 = s2.Length;
 
        if (canBeMapped(s1, l1, s2, l2))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code is contributed 
// by PrinciRaj1992 

Javascript

<script>
 
// Javascript implementation of the approach
 
var MAX = 26;
 
// Function that returns true if the mapping is possible
function canBeMapped(s1, l1, s2, l2)
{
 
    // Both the strings are of un-equal lengths
    if (l1 != l2)
        return false;
 
    // To store the frequencies of the
    // characters in both the string
    var freq1 = Array(MAX).fill(0);
    var freq2 = Array(MAX).fill(0);
 
    // Update frequencies of the characters
    for (var i = 0; i < l1; i++)
        freq1[s1[i].charCodeAt(0) - 'a'.charCodeAt(0)]++;
    for (var i = 0; i < l2; i++)
        freq2[s2[i].charCodeAt(0) - 'a'.charCodeAt(0)]++;
 
    // For every character of s1
    for (var i = 0; i < MAX; i++) {
 
        // If current character is
        // not present in s1
        if (freq1[i] == 0)
            continue;
        var found = false;
 
        // Find a character in s2 that has frequency
        // equal to the current character's
        // frequency in s1
        for (var j = 0; j < MAX; j++) {
 
            // If such character is found
            if (freq1[i] == freq2[j]) {
 
                // Set the frequency to -1 so that
                // it doesn't get picked again
                freq2[j] = -1;
 
                // Set found to true
                found = true;
                break;
            }
        }
 
        // If there is no character in s2
        // that could be mapped to the
        // current character in s1
        if (!found)
            return false;
    }
 
    return true;
}
 
// Driver code
var s1 = "axx";
var s2 = "cbc";
var l1 = s1.length;
var l2 = s2.length;
if (canBeMapped(s1, l1, s2, l2))
    document.write( "Yes");
else
    document.write( "No");
 
</script>

Output:

Yes

Time Complexity: O(26*(L1+L2))

Auxiliary Space: O(26)

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Map every character of one string to another such that all occurrences are mapped to the same character

C++

Java

Python3

C#

Javascript

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