Check whether the string S1 can be made equal to S2 with the given operation

Given two string S1 and S2, the task is to check whether both the strings can be made equal by performing the given operation on string S1. In a single operation, any character at an odd index can be swapped with any other character at an odd index, the same goes for the characters at even indices.
Examples:

Input: S1 = “abcd”, S2 = “cbad” 
Output: Yes 
Swap ‘a’ and ‘c’ in S1 and the resultant 
string is equal to S2.
Input: S1 = “abcd”, S2 = “abcdcd” 
Output: No 

Approach:

• Create a string even_s1 from the characters at even indices from S1.
• Similarly, generate the strings even_s2, odd_s1 and odd_s2.
• Sort all the four strings from the previous steps.
• If even_s1 = even_s2 and odd_s1 = odd_s2 then print Yes.
• Else print No as the strings cannot be made equal.

Below is the implementation of the above approach: 

C++

// CPP implementation of the approach
#include<bits/stdc++.h>
using namespace std;

// Function to return the string formed
// by the odd indexed characters of s
string partOdd(string s)
{
    string st = "";
    for(int i = 0; i < s.length(); i++)
    {
        if (i % 2 != 0)
        st += s[i];
    }
    return st;
}

// Function to return the string formed
// by the even indexed characters of s
string partEven(string str)
{
    string s = "";
    for(int i = 0; i < str.length(); i++)
    {
        if (i % 2 == 0)
        s += str[i];
    }
    return s;
}

// Function that returns true if s1
// can be made equal to s2
// with the given operation
bool canBeMadeEqual(string s1, string s2)
{

    // Get the string formed by the
    // even indexed characters of s1
    string even_s1 = partEven(s1);


    // Get the string formed by the
    // even indexed characters of s2
    string even_s2 = partEven(s2);

    // Get the string formed by the
    // odd indexed characters of s1
    string odd_s1 = partOdd(s1);

    // Get the string formed by the
    // odd indexed characters of s2
    string odd_s2 = partOdd(s2);

    // Sorting all the lists
    sort(even_s1.begin(), even_s1.end());
    sort(even_s2.begin(), even_s2.end());
    sort(odd_s1.begin(), odd_s1.end());
    sort(odd_s2.begin(), odd_s2.end());

    // If the strings can be made equal
    if (even_s1 == even_s2 and odd_s1 == odd_s2)
        return true;

    return false;
}

// Driver code
int main()
{
    string s1 = "cdab";
    string s2 = "abcd";
    if(canBeMadeEqual(s1, s2))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
}

// This code is contributed by Surendra_Gangwar

Java

// Java implementation of the approach
import java.util.*;

class GFG
{

    // Function to return the string formed
    // by the odd indexed characters of s
    static String partOdd(String s)
    {
        String st = "";
        for (int i = 0; i < s.length(); i++)
        {
            if (i % 2 != 0)
                st += s.charAt(i);
        }

        return st;
    }

    // Function to return the string formed
    // by the even indexed characters of s
    static String partEven(String str)
    {
        String s = "";
        for (int i = 0; i < str.length(); i++)
        {
            if (i % 2 == 0)
                s += str.charAt(i);
        }
        return s;
    }

    // Function that returns true if s1
    // can be made equal to s2
    // with the given operation
    static boolean canBeMadeEqual(String s1,
                                  String s2)
    {

        // Get the string formed by the
        // even indexed characters of s1
        char[] even1 = partEven(s1).toCharArray();

        // Get the string formed by the
        // even indexed characters of s2
        char[] even2 = partEven(s2).toCharArray();

        // Get the string formed by the
        // odd indexed characters of s1
        char[] odd1 = partOdd(s1).toCharArray();

        // Get the string formed by the
        // odd indexed characters of s2
        char[] odd2 = partOdd(s2).toCharArray();

        // Sorting all the lists
        Arrays.sort(even1);
        Arrays.sort(even2);
        Arrays.sort(odd1);
        Arrays.sort(odd2);

        String even_s1 = new String(even1);
        String even_s2 = new String(even2);
        String odd_s1 = new String(odd1);
        String odd_s2 = new String(odd2);

        // If the strings can be made equal
        if (even_s1.equals(even_s2) &&
             odd_s1.equals(odd_s2))
            return true;
        return false;
    }

    // Driver Code
    public static void main(String[] args)
    {
        String s1 = "cdab";
        String s2 = "abcd";

        if (canBeMadeEqual(s1, s2))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

// This code is contributed by
// sanjeev2552

Python3

# Python3 implementation of the approach

# Function to return the string formed
# by the odd indexed characters of s
def partOdd(s):
    odd = []
    for i in range(len(s)):
        if i % 2 != 0:
            odd.append(s[i])
    return odd

# Function to return the string formed
# by the even indexed characters of s
def partEven(s):
    even = []
    for i in range(len(s)):
        if i % 2 == 0:
            even.append(s[i])
    return even


# Function that returns true if s1
# can be made equal to s2
# with the given operation
def canBeMadeEqual(s1, s2):

    # Get the string formed by the
    # even indexed characters of s1
    even_s1 = partEven(s1)


    # Get the string formed by the
    # even indexed characters of s2
    even_s2 = partEven(s2)

    # Get the string formed by the
    # odd indexed characters of s1
    odd_s1 = partOdd(s1)

    # Get the string formed by the
    # odd indexed characters of s2
    odd_s2 = partOdd(s2)

    # Sorting all the lists
    even_s1.sort()
    even_s2.sort()
    odd_s1.sort()
    odd_s2.sort()

    # If the strings can be made equal
    if even_s1 == even_s2 and odd_s1 == odd_s2:
        return True

    return False


# Driver code
s1 = "cdab"
s2 = "abcd"
if canBeMadeEqual(s1, s2):
    print("Yes")
else:
    print("No")

C#

// C# implementation of the approach
using System;

class GFG
{

    // Function to return the string formed
    // by the odd indexed characters of s
    static string partOdd(string s)
    {
        string st = "";
        for (int i = 0; i < s.Length; i++)
        {
            if (i % 2 != 0)
                st += s[i];
        }

        return st;
    }

    // Function to return the string formed
    // by the even indexed characters of s
    static string partEven(string str)
    {
        string s = "";
        for (int i = 0; i < str.Length; i++)
        {
            if (i % 2 == 0)
                s += str[i];
        }
        return s;
    }

    // Function that returns true if s1
    // can be made equal to s2
    // with the given operation
    static bool canBeMadeEqual(string s1,
                                string s2)
    {

        // Get the string formed by the
        // even indexed characters of s1
        char[] even1 = partEven(s1).ToCharArray();

        // Get the string formed by the
        // even indexed characters of s2
        char[] even2 = partEven(s2).ToCharArray();

        // Get the string formed by the
        // odd indexed characters of s1
        char[] odd1 = partOdd(s1).ToCharArray();

        // Get the string formed by the
        // odd indexed characters of s2
        char[] odd2 = partOdd(s2).ToCharArray();

        // Sorting all the lists
        Array.Sort(even1);
        Array.Sort(even2);
        Array.Sort(odd1);
        Array.Sort(odd2);

        string even_s1 = new string(even1);
        string even_s2 = new string(even2);
        string odd_s1 = new string(odd1);
        string odd_s2 = new string(odd2);

        // If the strings can be made equal
        if (even_s1.Equals(even_s2) &&
            odd_s1.Equals(odd_s2))
            return true;
        return false;
    }

    // Driver Code
    public static void Main()
    {
        string s1 = "cdab";
        string s2 = "abcd";

        if (canBeMadeEqual(s1, s2))
            Console.Write("Yes");
        else
            Console.Write("No");
    }
}

// This code is contributed by AbhiThakur

Javascript

<script>

      // JavaScript implementation of the approach

      // Function to return the string formed
      // by the odd indexed characters of s
      function partOdd(s) {
        var st = "";
        for (var i = 0; i < s.length; i++) {
          if (i % 2 !== 0) st += s[i];
        }

        return st;
      }

      // Function to return the string formed
      // by the even indexed characters of s
      function partEven(str) {
        var s = "";
        for (var i = 0; i < str.length; i++) {
          if (i % 2 === 0) s += str[i];
        }
        return s;
      }

      // Function that returns true if s1
      // can be made equal to s2
      // with the given operation
      function canBeMadeEqual(s1, s2) {
        // Get the string formed by the
        // even indexed characters of s1
        var even1 = partEven(s1).split("");

        // Get the string formed by the
        // even indexed characters of s2
        var even2 = partEven(s2).split("");

        // Get the string formed by the
        // odd indexed characters of s1
        var odd1 = partOdd(s1).split("");

        // Get the string formed by the
        // odd indexed characters of s2
        var odd2 = partOdd(s2).split("");

        // Sorting all the lists
        even1.sort();
        even2.sort();
        odd1.sort();
        odd2.sort();

        var even_s1 = even1.join("");
        var even_s2 = even2.join("");
        var odd_s1 = odd1.join("");
        var odd_s2 = odd2.join("");

        // If the strings can be made equal
        if (even_s1 === even_s2 && odd_s1 === odd_s2) return true;
        return false;
      }

      // Driver Code
      var s1 = "cdab";
      var s2 = "abcd";

      if (canBeMadeEqual(s1, s2)) document.write("Yes");
      else document.write("No");

 </script>

Output:

Yes

Time Complexity: O(n*log(n)+m*log(m)) where n and m are the lengths of the string.
Auxiliary Space: O(n+m)