Check if two Binary Strings can be made equal by doing bitwise XOR of adjacent

Given binary strings S1 and S2 of length N, the task is to check if S2 can be made equal to S1 by performing the following operations on S2:

• The first operation is S_i = S_i ? S_i+1. ( ? is the XOR operation)
• The second operation is S_i+1 = S_i+1 ? S_i.

Examples:

Input: S1 = "00100", S2 = "00011" 
Output: Yes 
?Explanation: We can apply the following operations on S2 :
1. Select i = 2, So the conversion is 00011 ? 00111 
2. Select i = 3. So the conversion is 00111 ? 00100.
Hence it is possible to make S2 equal to S1 by applying the above operation on index 2 and 3.

Input: S1 = "10101", S2 = "01000" 
?Output: No

Approach: The problem can be solved based on the following observation: 

The operations have the following effects:

1. 01 becomes 11
2. 10 becomes 11
3. 00 becomes 00
4. 11 becomes 00

Let’s get the easy cases out of the way:

• If S2 = S1, the answer is YES
• If S2 is 000...00 (i.e S2 does not have a 1), then we can only perform the operation and it will be the third case shown above, which does not change anything. Therefore, the answer is NO if S2 ? S1.
• Now, assume S2 ? S1 and S2 has a 1. It is possible to convert S2 to S1 if and only if S1 has at least two consecutive same characters.

Follow the steps mentioned below to implement the above idea:

• Check if the strings are already equal to each other. If so, then return true.
• Otherwise, check if S2 has at least one '1':
  • If it has, then check for the condition that S1 has at least two consecutive same characters.
  • If the above condition is satisfied then return true.
  • Otherwise, return false.

Below is the implementation of the above approach.

// c++ code to implement the approach
#include <iostream>
using namespace std;

// Function to check whether S2 can be made equal
// to S1 after performing given operations
void checkEqual(string S1, string S2, int n)
{
    int c_1 = 0;
    int c_2 = 0;
    if (S2==S1) {
            cout << "Yes";
    }
    for (int j = 0; j < n; j++) {
        if (S2[j] == '1') {
            c_1++;
        }
        if (S1[j] == '1') {
            c_2++;
        }
    }
    if (c_1 == 0 && c_2 > 0) {
        cout << "No";
    }
    int c = 0;
    for (int k = 0; k < n - 1; k++) {
        if (S1[k] != S1[k + 1]) {
            c++;
        }
    }
    if (c == n - 1) {
        cout << "No";
    }
    else {
        cout << "Yes";
    }
}
int main()
{
    string S1 = "00100";
    string S2 = "00011";
    int N = S1.length();

    // Function call
    checkEqual(S1, S2, N);
    return 0;
}
//this code is contributed by aditya942003patil

// Java code to implement the approach

import java.io.*;
import java.util.*;

class GFG {

    // Function to check whether S2 can be made equal
    // to S1 after performing given operations
    public static void checkEqual(String S1,
                                  String S2, int n)
    {
        char[] a = S2.toCharArray();
        char[] b = S1.toCharArray();

        int c_1 = 0;
        int c_2 = 0;
        if (String.valueOf(a).equals(String.valueOf(b))) {
            System.out.println("Yes");
        }
        for (int j = 0; j < n; j++) {
            if (a[j] == '1') {
                c_1++;
            }
            if (b[j] == '1') {
                c_2++;
            }
        }
        if (c_1 == 0 && c_2 > 0) {
            System.out.println("No");
        }
        int c = 0;
        for (int k = 0; k < n - 1; k++) {
            if (b[k] != b[k + 1]) {
                c++;
            }
        }
        if (c == n - 1) {
            System.out.println("No");
        }
        else {
            System.out.println("Yes");
        }
    }

    // Driver code
    public static void main(String[] args)
    {
        String S1 = "00100";
        String S2 = "00011";
        int N = S1.length();

        // Function call
        checkEqual(S1, S2, N);
    }
}

# python code to implement the approach
# Function to check whether S2 can be made equal
# to S1 after performing given operations
def checkEqual(S1, S2, n):
  c_1 = 0
  c_2 = 0
  if (S2 == S1):
    print("yes")
  for j in range(n):
    if (S2[j] == '1'):
      c_1 += 1
    if (S1[j] == '1'):
      c_2 += 1
  if (c_1 == 0 and c_2 > 0):
    print("No")
  c = 0
  for k in range(n-1):
    if (S1[k] != S1[k + 1]):
      c += 1
  if (c == n - 1):
    print("No")
  else:
    print("yes")
S1 = "00100"
S2 = "00011"
N = len(S1)

# Function call
checkEqual(S1, S2, N);

# This code is contributed by Atul_kumar_Shrivastava

// C# code to implement the approach


using System;

public class GFG {

    // Function to check whether S2 can be made equal
    // to S1 after performing given operations
    public static void checkEqual(String S1,
                                  String S2, int n)
    {
        char[] a = S2.ToCharArray();
        char[] b = S1.ToCharArray();

        int c_1 = 0;
        int c_2 = 0;
        if (String.Join("",a).Equals(String.Join("",b))) {
            Console.WriteLine("Yes");
        }
        for (int j = 0; j < n; j++) {
            if (a[j] == '1') {
                c_1++;
            }
            if (b[j] == '1') {
                c_2++;
            }
        }
        if (c_1 == 0 && c_2 > 0) {
            Console.WriteLine("No");
        }
        int c = 0;
        for (int k = 0; k < n - 1; k++) {
            if (b[k] != b[k + 1]) {
                c++;
            }
        }
        if (c == n - 1) {
            Console.WriteLine("No");
        }
        else {
            Console.WriteLine("Yes");
        }
    }

    // Driver code
    public static void Main(String[] args)
    {
        String S1 = "00100";
        String S2 = "00011";
        int N = S1.Length;

        // Function call
        checkEqual(S1, S2, N);
    }
}

// This code contributed by shikhasingrajput 

<script>
    // JavaScript program of above approach.

    // Function to check whether S2 can be made equal
    // to S1 after performing given operations
    function checkEqual(S1, S2, n)
    {
        let a = S2.split();
        let b = S1.split();

        let c_1 = 0;
        let c_2 = 0;
        if (a == b) {
            document.write("Yes");
        }
        for (let j = 0; j < n; j++) {
            if (a[j] == '1') {
                c_1++;
            }
            if (b[j] == '1') {
                c_2++;
            }
        }
        if (c_1 == 0 && c_2 > 0) {
            document.write("No");
        }
        let c = 0;
        for (let k = 0; k < n - 1; k++) {
            if (b[k] != b[k + 1]) {
                c++;
            }
        }
        if (c == n - 1) {
            document.write("No");
        }
        else {
            document.write("Yes");
        }
    }

    // Driver code

        let S1 = "00100";
        let S2 = "00011";
        let N = S1.length;

        // Function call
        checkEqual(S1, S2, N);

// This code is contributed by sanjoy_62.
</script>

Yes

Time Complexity: O(N) 
Auxiliary Space: O(1)