Check whether two strings can be made equal by copying their characters with the adjacent ones

Given two strings str1 and str2, the task is to check whether both of the string can be made equal by copying any character of the string with its adjacent character. Note that this operation can be performed any number of times.
Examples:

Input: str1 = “abc”, str2 = “def” 
Output: No 
As all the characters in both the string are different. 
So, there is no way they can be made equal.
Input: str1 = “abc”, str2 = “fac” 
Output: Yes 
str1 = “abc” -> “aac” 
str2 = “fac” -> “aac” 

Approach: In order for the strings to be made equal with the given operation, they have to be of equal lengths and there has to be at least one character which is common in both the strings. To check that, create a frequency array freq[] which will store the frequency of all the characters of str1 and then for every character of str2 if its frequency in str1 is greater than 0 then it is possible to make both the strings equal.
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 both
// the strings can be made equal
// with the given operation
bool canBeMadeEqual(string str1, string str2)
{
    int len1 = str1.length();
    int len2 = str2.length();

    // Lengths of both the strings
    // have to be equal
    if (len1 == len2) {

        // To store the frequency of the
        // characters of str1
        int freq[MAX];
        for (int i = 0; i < len1; i++) {
            freq[str1[i] - 'a']++;
        }

        // For every character of str2
        for (int i = 0; i < len2; i++) {

            // If current character of str2
            // also appears in str1
            if (freq[str2[i] - 'a'] > 0)
                return true;
        }
    }

    return false;
}

// Driver code
int main()
{
    string str1 = "abc", str2 = "defa";

    if (canBeMadeEqual(str1, str2))
        cout << "Yes";
    else
        cout << "No";

    return 0;
}

Java

// Java implementation of the above approach
class GFG
{
    static int MAX = 26;

    // Function that returns true if both
    // the strings can be made equal
    // with the given operation
    static boolean canBeMadeEqual(String str1,
                                  String str2)
    {
        int len1 = str1.length();
        int len2 = str2.length();

        // Lengths of both the strings
        // have to be equal
        if (len1 == len2)
        {

            // To store the frequency of the
            // characters of str1
            int freq[] = new int[MAX];

            for (int i = 0; i < len1; i++)
            {
                freq[str1.charAt(i) - 'a']++;
            }

            // For every character of str2
            for (int i = 0; i < len2; i++)
            {

                // If current character of str2
                // also appears in str1
                if (freq[str2.charAt(i) - 'a'] > 0)
                    return true;
            }
        }
        return false;
    }

    // Driver code
    public static void main (String[] args)
    {
        String str1 = "abc", str2 = "defa";

        if (canBeMadeEqual(str1, str2))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
MAX = 26

# Function that returns true if both
# the strings can be made equal
# with the given operation
def canBeMadeEqual(str1, str2):
    len1 = len(str1)
    len2 = len(str2)

    # Lengths of both the strings
    # have to be equal
    if (len1 == len2):

        # To store the frequency of the
        # characters of str1
        freq = [0 for i in range(MAX)]
        for i in range(len1):
            freq[ord(str1[i]) - ord('a')] += 1

        # For every character of str2
        for i in range(len2):

            # If current character of str2
            # also appears in str1
            if (freq[ord(str2[i]) - ord('a')] > 0):
                return True

    return False

# Driver code
str1 = "abc"
str2 = "defa"

if (canBeMadeEqual(str1, str2)):
    print("Yes")
else:
    print("No")

# This code is contributed by Mohit Kumar

C#

// C# implementation of the above approach
using System;

class GFG
{
    static int MAX = 26;

    // Function that returns true if both
    // the strings can be made equal
    // with the given operation
    static Boolean canBeMadeEqual(String str1,
                                   String str2)
    {
        int len1 = str1.Length;
        int len2 = str2.Length;

        // Lengths of both the strings
        // have to be equal
        if (len1 == len2)
        {

            // To store the frequency of the
            // characters of str1
            int []freq = new int[MAX];

            for (int i = 0; i < len1; i++)
            {
                freq[str1[i] - 'a']++;
            }

            // For every character of str2
            for (int i = 0; i < len2; i++)
            {

                // If current character of str2
                // also appears in str1
                if (freq[str2[i] - 'a'] > 0)
                    return true;
            }
        }
        return false;
    }

    // Driver code
    public static void Main (String []args)
    {
        String str1 = "abc", str2 = "defa";

        if (canBeMadeEqual(str1, str2))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

// This code is contributed by Arnab Kundu

Javascript

<script>

    // JavaScript implementation of the above approach

    let MAX = 26;

    // Function that returns true if both
    // the strings can be made equal
    // with the given operation
    function canBeMadeEqual(str1, str2)
    {
        let len1 = str1.length;
        let len2 = str2.length;

        // Lengths of both the strings
        // have to be equal
        if (len1 == len2)
        {

            // To store the frequency of the
            // characters of str1
            let freq = new Array(MAX);
            freq.fill(0);

            for (let i = 0; i < len1; i++)
            {
                freq[str1[i].charCodeAt() -
                'a'.charCodeAt()]++;
            }

            // For every character of str2
            for (let i = 0; i < len2; i++)
            {

                // If current character of str2
                // also appears in str1
                if (freq[str2[i].charCodeAt() -
                'a'.charCodeAt()] > 0)
                    return true;
            }
        }
        return false;
    }

    let str1 = "abc", str2 = "defa";

    if (canBeMadeEqual(str1, str2))
      document.write("Yes");
    else
      document.write("No");

</script>

Output:

No

Time Complexity: O(N)

Auxiliary Space: O(26)