Check if two binary strings can be made equal by swapping 1s occurring before 0s

Given two binary strings str1 and str2 having same length, the task is to find if it is possible to make the two binary strings str1 and str2 equal by swapping all 1s occurring at indices less than that of 0s index in the binary string str1.

Examples:

Input: str1 = “0110”, str2 = “0011”
Output: Possible
Explanation:
Swapping str1[2] with str1[3] the binary string str1 becomes “0101”.
Swapping str1[1] with str1[2] the binary string str1 becomes “0011” .
The binary string str1 becomes equal to the binary string str2 therefore, the required output is Possible.
Input: str1 = “101”, str2 = “010”
Output: Not Possible

Approach: The idea is to count the number of 1s and 0s in str1 and str2 and then proceed accordingly. Follow the steps below to solve the problem:

If the count of 1s and 0s are not equal in str1 and str2, then conversion is not possible.
Traverse the string.
Starting from the first character, compare each character one by one. For each different character at i, perform the following steps:
- Check if the current character of the string str1 is ‘0’ and curStr1Ones (stores the current count of 1’s of the string str1) is greater than 0. If found to be true, then swap the character with ‘1’ and decrement the value of curStr1Ones by 1.
- Check if the character of the string str1 is ‘0’ and curStr1Ones is equal to 0. If found to be true, then increment the value of the flag by 1 and break the loop.
- Check if the character of the string str1 is ‘1’ and the character of the string str2 is ‘0’. If found to be true, then swap the character of str1 with ‘0’ and increment the value of curStr1Ones by 1.
Finally, print “Possible” if the flag is 0, otherwise print “Not Possible”.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if it is possible to make
// two binary strings equal by given operations
void isBinaryStringsEqual(string str1, string str2)
{
 
    // Stores count of 1's and 0's
    // of the string str1
    int str1Zeros = 0, str1Ones = 0;
 
    // Stores count of 1's and 0's
    // of the string str2
    int str2Zeros = 0, str2Ones = 0;
 
    int flag = 0;
 
    // Stores current count of 1's
    // present in the string str1
    int curStr1Ones = 0;
 
    // Count the number of 1's and 0's
    // present in the strings str1 and str2
    for (int i = 0; i < str1.length(); i++) {
 
        if (str1[i] == '1') {
            str1Ones++;
        }
 
        else if (str1[i] == '0') {
            str1Zeros++;
        }
 
        if (str2[i] == '1') {
            str2Ones++;
        }
 
        else if (str2[i] == '0') {
            str2Zeros++;
        }
    }
 
    // If the number of 1's and 0's
    // are not same of the strings str1
    // and str2 then print not possible
    if (str1Zeros != str2Zeros && str1Ones != str2Ones) {
        cout << "Not Possible";
    }
 
    else {
 
        // Traversing through the
        // strings str1 and str2
        for (int i = 0; i < str1.length(); i++) {
 
            // If the str1 character not
            // equals to str2 character
            if (str1[i] != str2[i]) {
 
                // Swaps 0 with 1 of the
                // string str1
                if (str1[i] == '0' && curStr1Ones > 0) {
                    str1[i] = '1';
                    curStr1Ones--;
                }
 
                // Breaks the loop as the count
                // of 1's is zero. Hence, no swaps possible
                if (str1[i] == '0' && curStr1Ones == 0) {
 
                    flag++;
                    break;
                }
 
                // Swaps 1 with 0 in the string str1
                if (str1[i] == '1' && str2[i] == '0') {
 
                    str1[i] = '0';
                    curStr1Ones++;
                }
            }
        }
 
        if (flag == 0) {
            cout << "Possible";
        }
 
        // Print not possible
        else {
            cout << "Not Possible";
        }
    }
}
 
// Driver Code
int main()
{
    // Given Strings
    string str1 = "0110";
    string str2 = "0011";
 
    // Function Call
    isBinaryStringsEqual(str1, str2);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
class GFG {
 
  // Function to check if it is possible to make
  // two binary strings equal by given operations
  static void isBinaryStringsEqual(String str1,
                                   String str2)
  {
 
    // Stores count of 1's and 0's
    // of the string str1
    int str1Zeros = 0, str1Ones = 0;
 
    // Stores count of 1's and 0's
    // of the string str2
    int str2Zeros = 0, str2Ones = 0;
    int flag = 0;
 
    // Stores current count of 1's
    // present in the string str1
    int curStr1Ones = 0;
 
    // Count the number of 1's and 0's
    // present in the strings str1 and str2
    for (int i = 0; i < str1.length(); i++)
    {
 
      if (str1.charAt(i) == '1')
      {
        str1Ones++;
      }
 
      else if (str1.charAt(i) == '0')
      {
        str1Zeros++;
      }
 
      if (str2.charAt(i) == '1')
      {
        str2Ones++;
      }
 
      else if (str2.charAt(i) == '0')
      {
        str2Zeros++;
      }
    }
 
    // If the number of 1's and 0's
    // are not same of the strings str1
    // and str2 then print not possible
    if (str1Zeros != str2Zeros
        && str1Ones != str2Ones)
    {
      System.out.println("Not Possible");
    }
 
    else {
 
      // Traversing through the
      // strings str1 and str2
      for (int i = 0; i < str1.length(); i++)
      {
 
        // If the str1 character not
        // equals to str2 character
        if (str1.charAt(i) != str2.charAt(i))
        {
 
          // Swaps 0 with 1 of the
          // string str1
          if (str1.charAt(i) == '0'
              && curStr1Ones > 0)
          {
            str1 = str1.substring(0, i) + '1'
              + str1.substring(i + 1);
            curStr1Ones--;
          }
 
          // Breaks the loop as the count
          // of 1's is zero. Hence, no swaps
          // possible
          if (str1.charAt(i) == '0'
              && curStr1Ones == 0)
          {
            flag++;
            break;
          }
 
          // Swaps 1 with 0 in the string str1
          if (str1.charAt(i) == '1'
              && str2.charAt(i) == '0')
          {
            str1 = str1.substring(0, i) + '0'
              + str1.substring(i+1);
            curStr1Ones++;
          }
        }
      }
 
      if (flag == 0) {
        System.out.println("Possible");
      }
 
      // Print not possible
      else {
        System.out.println("Not Possible");
      }
    }
  }
 
  // Driver Code
  public static void main(String[] args)
  {
 
    // Given Strings
    String str1 = "0110";
    String str2 = "0011";
 
    // Function Call
    isBinaryStringsEqual(str1, str2);
  }
}
 
// This code is contributed by dharanendralv23

Python3

# Python program for the above approach
 
# Function to check if it is possible to make
# two binary strings equal by given operations
def isBinaryStringsEqual(list1, list2) :
     
    str1 = list(list1)
    str2 = list(list2)
 
    # Stores count of 1's and 0's
    # of the string str1
    str1Zeros = 0
    str1Ones = 0
 
    # Stores count of 1's and 0's
    # of the string str2
    str2Zeros = 0
    str2Ones = 0
    flag = 0
 
    # Stores current count of 1's
    # present in the string str1
    curStr1Ones = 0
 
    # Count the number of 1's and 0's
    # present in the strings str1 and str2
    for i in range(len(str1)):
        if (str1[i] == '1') :
            str1Ones += 1
        elif (str1[i] == '0') :
            str1Zeros += 1
        if (str2[i] == '1') :
            str2Ones += 1
        elif (str2[i] == '0') :
            str2Zeros += 1
        
    # If the number of 1's and 0's
    # are not same of the strings str1
    # and str2 then print not possible
    if (str1Zeros != str2Zeros and str1Ones != str2Ones) :
        print("Not Possible")
    else :
 
        # Traversing through the
        # strings str1 and str2
        for i in range(len(str1)):
 
            # If the str1 character not
            # equals to str2 character
            if (str1[i] != str2[i]) :
 
                # Swaps 0 with 1 of the
                # string str1
                if (str1[i] == '0' and curStr1Ones > 0) :              
                    str1[i] = '1'
                    curStr1Ones -= 1
                 
                # Breaks the loop as the count
                # of 1's is zero. Hence, no swaps possible
                if (str1[i] == '0' and curStr1Ones == 0) :
                    flag += 1
                    break
                 
                # Swaps 1 with 0 in the string str1
                if (str1[i] == '1' and str2[i] == '0') :
                    str1[i] = '0'
                    curStr1Ones += 1
                 
        if (flag == 0) :
            print("Possible")
         
        # Print not possible
        else :
           print("Not Possible")
 
# Driver Code
 
# Given Strings
str1 = "0110"
str2 = "0011"
 
# Function Call
isBinaryStringsEqual(str1, str2)
 
# This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
using System.Text;
class GFG
{
 
  // Function to check if it is possible to make
  // two binary strings equal by given operations
  static void isBinaryStringsEqual(string str1,
                                   string str2)
  {
 
    // Stores count of 1's and 0's
    // of the string str1
    int str1Zeros = 0, str1Ones = 0;
 
    // Stores count of 1's and 0's
    // of the string str2
    int str2Zeros = 0, str2Ones = 0;
    int flag = 0;
 
    // Stores current count of 1's
    // present in the string str1
    int curStr1Ones = 0;
 
    // Count the number of 1's and 0's
    // present in the strings str1 and str2
    for (int i = 0; i < str1.Length; i++)
    {
      if (str1[i] == '1')
      {
        str1Ones++;
      }
 
      else if (str1[i] == '0')
      {
        str1Zeros++;
      }
 
      if (str2[i] == '1')
      {
        str2Ones++;
      }
 
      else if (str2[i] == '0')
      {
        str2Zeros++;
      }
    }
 
    // If the number of 1's and 0's
    // are not same of the strings str1
    // and str2 then print not possible
    if (str1Zeros != str2Zeros
        && str1Ones != str2Ones)
    {
      Console.WriteLine("Not Possible");
    }
 
    else
    {
 
      // Traversing through the
      // strings str1 and str2
      for (int i = 0; i < str1.Length; i++)
      {
 
        // If the str1 character not
        // equals to str2 character
        if (str1[i] != str2[i])
        {
 
          // Swaps 0 with 1 of the
          // string str1
          if (str1[i] == '0' && curStr1Ones > 0)
          {
            StringBuilder sb
              = new StringBuilder(str1);
            sb[i] = '1';
            str1 = sb.ToString();
            curStr1Ones--;
          }
 
          // Breaks the loop as the count
          // of 1's is zero. Hence, no swaps
          // possible
          if (str1[i] == '0'
              && curStr1Ones == 0)
          {
            flag++;
            break;
          }
 
          // Swaps 1 with 0 in the string str1
          if (str1[i] == '1' && str2[i] == '0')
          {
            StringBuilder sb
              = new StringBuilder(str1);
            sb[i] = '0';
            str1 = sb.ToString();
            curStr1Ones++;
          }
        }
      }
 
      if (flag == 0)
      {
        Console.WriteLine("Possible");
      }
 
      // Print not possible
      else
      {
        Console.WriteLine("Not Possible");
      }
    }
  }
 
  // Driver Code
  static public void Main()
  {
 
    // Given Strings
    string str1 = "0110";
    string str2 = "0011";
 
    // Function Call
    isBinaryStringsEqual(str1, str2);
  }
}
 
// This code is contributed by dharanendralv23

Javascript

<script>
 
      // JavaScript program for the above approach
       
      // Function to check if it is possible to make
      // two binary strings equal by given operations
      function isBinaryStringsEqual(list1, list2) {
        var str1 = list1.split("");
        var str2 = list2.split("");
        // Stores count of 1's and 0's
        // of the string str1
        var str1Zeros = 0,
          str1Ones = 0;
 
        // Stores count of 1's and 0's
        // of the string str2
        var str2Zeros = 0,
          str2Ones = 0;
        var flag = 0;
 
        // Stores current count of 1's
        // present in the string str1
        var curStr1Ones = 0;
 
        // Count the number of 1's and 0's
        // present in the strings str1 and str2
        for (var i = 0; i < str1.length; i++) {
          if (str1[i] === "1") {
            str1Ones++;
          } else if (str1[i] === "0") {
            str1Zeros++;
          }
 
          if (str2[i] === "1") {
            str2Ones++;
          } else if (str2[i] === "0") {
            str2Zeros++;
          }
        }
 
        // If the number of 1's and 0's
        // are not same of the strings str1
        // and str2 then print not possible
        if (str1Zeros !== str2Zeros && str1Ones !== str2Ones) {
          document.write("Not Possible");
        } else {
          // Traversing through the
          // strings str1 and str2
          for (var i = 0; i < str1.length; i++) {
            // If the str1 character not
            // equals to str2 character
            if (str1[i] !== str2[i]) {
              // Swaps 0 with 1 of the
              // string str1
              if (str1[i] === "0" && curStr1Ones > 0) {
                str1[i] = "1";
 
                curStr1Ones--;
              }
 
              // Breaks the loop as the count
              // of 1's is zero. Hence, no swaps
              // possible
              if (str1[i] === "0" && curStr1Ones === 0) {
                flag++;
                break;
              }
 
              // Swaps 1 with 0 in the string str1
              if (str1[i] === "1" && str2[i] === "0") {
                str1[i] = "0";
                curStr1Ones++;
              }
            }
          }
 
          if (flag === 0) {
            document.write("Possible");
          }
 
          // Print not possible
          else {
            document.write("Not Possible");
          }
        }
      }
 
      // Driver Code
      // Given Strings
      var str1 = "0110";
      var str2 = "0011";
 
      // Function Call
      isBinaryStringsEqual(str1, str2);
       
</script>

Output

Possible

Time Complexity: O(|str1|), where |str1| is the length of the input string. This is because the program loops through the strings once to count the number of ones and zeros, and then loops through the strings again to perform the swaps. The inner loop has a constant time complexity, so the overall time complexity is linear.
Auxiliary Space: O(1), The space complexity of the program is O(1), as it only uses a constant amount of extra space to store the counts and the flag variable. The input strings are not modified and do not take extra space as they are passed by value.

Another Approach:

Define a function named “areBinaryStringsEqual” that takes two string arguments: str1 and str2.
Initialize two integer variables ones1 and ones2 to count the number of 1s in both strings.
Traverse both strings using a for loop and count the number of 1s in each string.
If the number of 1s is not the same in both strings, return false.
Initialize a variable count to keep track of the number of 0s that need to be swapped in string 1.
Traverse both strings using a for loop and check if they can be made equal by swapping 1s and 0s.
If the characters at the same position in both strings are not equal, check the following:
a. If the character in string 1 is 1 and in string 2 is 0, increment the count.
b. If the character in string 1 is 0 and count is greater than 0, decrement the count.
c. If the characters cannot be made equal by swapping, return false.
If the loop completes without returning false, the strings can be made equal by swapping 1s and 0s. Return true.
In the main function, input two binary strings str1 and str2.
Check if the strings can be made equal by calling the “areBinaryStringsEqual” function.
If the function returns true, print “Possible”. Otherwise, print “Not Possible”.
Return 0 to end the program.

Below is the implementation of the above approach:

C++

// Include necessary header files
#include <iostream>
#include <string>
using namespace std;
 
// Function to check if two binary strings can be made equal by swapping 1s and 0s
bool areBinaryStringsEqual(string str1, string str2) {
// Count number of 1s in both strings
int ones1 = 0, ones2 = 0;
for (int i = 0; i < str1.length(); i++) {
if (str1[i] == '1') ones1++;
if (str2[i] == '1') ones2++;
}
 
 
// If the number of 1s is not the same in both strings, return false
if (ones1 != ones2) return false;
 
// Count variable to keep track of number of 0s that need to be swapped in string 1
int count = 0;
 
// Traverse both strings and check if they can be made equal by swapping 1s and 0s
for (int i = 0; i < str1.length(); i++) {
    // If the characters at the same position in both strings are not equal
    if (str1[i] != str2[i]) {
        // If the character in string 1 is 1 and in string 2 is 0, increment the count
        if (str1[i] == '1' && str2[i] == '0') count++;
        // If the character in string 1 is 0 and count is greater than 0, decrement the count
        else if (str1[i] == '0' && count > 0) count--;
        // If the characters cannot be made equal by swapping, return false
        else return false;
    }
}
 
// If the loop completes without returning false, the strings can be made equal by swapping 1s and 0s
return true;
}
 
// Driver code
int main() {
// Input binary strings
string str1 = "0110";
string str2 = "0011";
 
 
// Check if the strings can be made equal and print output accordingly
if (areBinaryStringsEqual(str1, str2)) {
    cout << "Possible";
} else {
    cout << "Not Possible";
}
 
// Return 0 to end the program
return 0;
}
//This code is contributed by rudra1807raj

Java

public class Main {
    // Function to check if two binary
    // strings can be made equal by swapping 1s and 0s
    public static boolean areBinaryStringsEqual(String str1, String str2) {
        // Count number of 1s in both strings
        int ones1 = 0, ones2 = 0;
        for (int i = 0; i < str1.length(); i++) {
            if (str1.charAt(i) == '1') ones1++;
            if (str2.charAt(i) == '1') ones2++;
        }
 
        // If the number of 1s is not the same in both strings, return false
        if (ones1 != ones2) return false;
 
        // Count variable to keep track of
        // number of 0s that need to be swapped in string 1
        int count = 0;
 
        // Traverse both strings and check if
        // they can be made equal by swapping 1s and 0s
        for (int i = 0; i < str1.length(); i++) {
            // If the characters at the same position
            // in both strings are not equal
            if (str1.charAt(i) != str2.charAt(i)) {
                // If the character in string 1
                // is 1 and in string 2 is 0, increment the count
                if (str1.charAt(i) == '1' && str2.charAt(i) == '0') count++;
                // If the character in string 1 is 0
                // and count is greater than 0, decrement the count
                else if (str1.charAt(i) == '0' && count > 0) count--;
                // If the characters cannot be made equal by swapping, return false
                else return false;
            }
        }
 
        // If the loop completes without returning false,
        // the strings can be made equal by swapping 1s and 0s
        return true;
    }
 
    // Driver code
    public static void main(String[] args) {
        // Input binary strings
        String str1 = "0110";
        String str2 = "0011";
 
        // Check if the strings can be
        // made equal and print output accordingly
        if (areBinaryStringsEqual(str1, str2)) {
            System.out.println("Possible");
        } else {
            System.out.println("Not Possible");
        }
    }
}

Output

Possible

Time complexity: The code traverses each character of the input strings twice, so the time complexity is O(n), where n is the length of the strings.

Auxiliary Space: The code uses a constant amount of extra space, so the space complexity is O(1).

Last Updated : 13 Sep, 2023