# 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”

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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

Output:

No

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