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

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)