Minimum prefixes required to be flipped to convert a Binary String to another

Given two binary strings A and B of length N, the task is to convert the string from A to string B by repeatedly flipping all the bits of a prefix of A, i.e. convert all the 0s in the prefix to 1s and vice-versa and print the minimum number of prefix flips required and the length of respective prefixes.

Examples:

Input: A = "001", B = "000"
Output: 
2
3 2
Explanation: 
Flipping the prefix "001" modifies the string to "110".
Flipping the prefix "11" modifies the string to "000".

Input: A = "1000", B = "1011"
Output: 
2
4 2
Explanation: 
Flipping the prefix "1000" modifies the string to "0111".
Flipping the prefix "01" modifies the string to "1011".

Naive Approach: The simplest approach is to traverse the string A in reverse and for every i^th character obtained such that A[i] is not equal to B[i], flip the characters present in A from indices [0, i] and increment the number of operations by 1. After complete traversal of the string, print the count of operations and the prefix length chosen in each operation.
Time Complexity: O(N²)
Auxiliary Space: O(N)

Efficient Approach: The idea is to fix one bit at a time by traversing the string A in reverse. Maintain a boolean variable, say invert, initially set to false, to denote whether the current bits in A are flipped or not. While traversing, perform the following:

• If the iᵗʰ bit in A is not equal to the iᵗʰ bit in B and invert is false, then increment the count of operations and set invert to true.
• Otherwise, if the iᵗʰ bit in A is equal to the iᵗʰ bit in B and invert is true, then increment the count of operations and set invert to false.

Follow the steps below to solve the problem:

• Initialize a boolean variable, say invert as false, to denote whether the bits in A are flipped or not.
• Initialize an empty array, say res, to store the prefix length in each operation.
• Iterate in the range [N - 1, 0] using the variable i and perform the following steps:
  • If A[i] != B[i] and invert is false, then the current bit is required to be flipped. Therefore. insert (i + 1) into the array res and update invert to true.
  • Otherwise, check if A[i] == B[i] and invert is true, then insert (i + 1) to res, and update invert to false.
• Print the size of the array res as the number of operations required to make the two strings equal.
• Then, print the values stored in res to denote the prefix length in each operation.

Below is the implementation of the above approach:

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to count flips required
// to make strings A and B equal
void findOperations(string A,
                    string B, int N)
{
    // Stores the count of the
    // number of operations
    int operations = 0;

    // Stores the length of
    // the chosen prefixes
    vector<int> ops;

    // Stores if operations are
    // performed even or odd times
    bool invert = false;

    // Traverse the given string
    for (int i = N - 1; i >= 0; i--) {

        // If current characters in
        // the two strings are unequal
        if (A[i] != B[i]) {

            // If A[i] is not flipped
            if (!invert) {

                // Increment count
                // of operations
                operations++;

                // Insert the length of
                // the chosen prefix
                ops.push_back(i + 1);

                // Update invert to true
                invert = true;
            }
        }

        else {

            // If A[i] is flipped
            if (invert) {

                // Increment count
                // of operations
                operations++;

                // Insert length of
                // the chosen prefix
                ops.push_back(i + 1);

                // Update invert to false
                invert = false;
            }
        }
    }

    // Print the number of
    // operations required
    cout << operations << endl;

    // Print the chosen prefix
    // length in each operation
    if (operations != 0) {
        for (auto x : ops)
            cout << x << " ";
    }
}

// Driver Code
int main()
{
    // Given binary strings
    string A = "001", B = "000";
    int N = A.size();

    findOperations(A, B, N);

    return 0;
}

// Java program for the above approach

import java.util.*;

class GFG{

// Function to count flips required
// to make Strings A and B equal
static void findOperations(String A,
                    String B, int N)
{
    // Stores the count of the
    // number of operations
    int operations = 0;

    // Stores the length of
    // the chosen prefixes
    Vector<Integer> ops =  new Vector<>();

    // Stores if operations are
    // performed even or odd times
    boolean invert = false;

    // Traverse the given String
    for (int i = N - 1; i >= 0; i--) {

        // If current characters in
        // the two Strings are unequal
        if (A.charAt(i) != B.charAt(i)) {

            // If A[i] is not flipped
            if (!invert) {

                // Increment count
                // of operations
                operations++;

                // Insert the length of
                // the chosen prefix
                ops.add(i + 1);

                // Update invert to true
                invert = true;
            }
        }

        else {

            // If A[i] is flipped
            if (invert) {

                // Increment count
                // of operations
                operations++;

                // Insert length of
                // the chosen prefix
                ops.add(i + 1);

                // Update invert to false
                invert = false;
            }
        }
    }

    // Print the number of
    // operations required
    System.out.print(operations +"\n");

    // Print the chosen prefix
    // length in each operation
    if (operations != 0) {
        for (int x : ops)
            System.out.print(x+ " ");
    }
}

// Driver Code
public static void main(String[] args)
{
    // Given binary Strings
    String A = "001", B = "000";
    int N = A.length();

    findOperations(A, B, N);

}
}

// This code is contributed by Amit Katiyar

# Python program for the above approach

# Function to count flips required
# to make strings A and B equal
def findOperations(A, B, N):
  
    # Stores the count of the
    # number of operations
    operations = 0

    # Stores the length of
    # the chosen prefixes
    ops = []

    # Stores if operations are
    # performed even or odd times
    invert = False

    # Traverse the given string
    for i in range(N - 1, -1, -1):
      
        # If current characters in
        # the two strings are unequal
        if (A[i] != B[i]):

            # If A[i] is not flipped
            if (not invert):

                # Increment count
                # of operations
                operations += 1

                # Insert the length of
                # the chosen prefix
                ops.append(i + 1)

                # Update invert to true
                invert = True
        else:

            # If A[i] is flipped
            if (invert):

                # Increment count
                # of operations
                operations += 1

                # Insert length of
                # the chosen prefix
                ops.append(i + 1)

                # Update invert to false
                invert = False

    # Print the number of
    # operations required
    print (operations)

    # Print the chosen prefix
    # length in each operation
    if (operations != 0):
        for x in ops:
            print(x, end = " ")

# Driver Code
if __name__ == '__main__':
  
    # Given binary strings
    A, B = "001", "000"
    N = len(A)

    findOperations(A, B, N)

# This code is contributed by mohit kumar 29.

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function to count flips required
// to make Strings A and B equal
static void findOperations(String A,
                           String B, int N)
{
    
    // Stores the count of the
    // number of operations
    int operations = 0;

    // Stores the length of
    // the chosen prefixes
    List<int> ops =  new List<int>();

    // Stores if operations are
    // performed even or odd times
    bool invert = false;

    // Traverse the given String
    for(int i = N - 1; i >= 0; i--) 
    {
        
        // If current characters in
        // the two Strings are unequal
        if (A[i] != B[i]) 
        {
            
            // If A[i] is not flipped
            if (!invert)
            {
                
                // Increment count
                // of operations
                operations++;

                // Insert the length of
                // the chosen prefix
                ops.Add(i + 1);

                // Update invert to true
                invert = true;
            }
        }

        else
        {
            
            // If A[i] is flipped
            if (invert) 
            {
                
                // Increment count
                // of operations
                operations++;

                // Insert length of
                // the chosen prefix
                ops.Add(i + 1);

                // Update invert to false
                invert = false;
            }
        }
    }

    // Print the number of
    // operations required
    Console.Write(operations + "\n");

    // Print the chosen prefix
    // length in each operation
    if (operations != 0) 
    {
        foreach(int x in ops)
            Console.Write(x + " ");
    }
}

// Driver Code
public static void Main(String[] args)
{
    
    // Given binary Strings
    String A = "001", B = "000";
    int N = A.Length;

    findOperations(A, B, N);
}
}

// This code is contributed by 29AjayKumar 

<script>
      // JavaScript program for the above approach
      // Function to count flips required
      // to make Strings A and B equal
      function findOperations(A, B, N) {
        // Stores the count of the
        // number of operations
        var operations = 0;

        // Stores the length of
        // the chosen prefixes
        var ops = [];

        // Stores if operations are
        // performed even or odd times
        var invert = false;

        // Traverse the given String
        for (var i = N - 1; i >= 0; i--) {
          // If current characters in
          // the two Strings are unequal
          if (A[i] !== B[i]) {
            // If A[i] is not flipped
            if (!invert) {
              // Increment count
              // of operations
              operations++;

              // Insert the length of
              // the chosen prefix
              ops.push(i + 1);

              // Update invert to true
              invert = true;
            }
          } else {
            // If A[i] is flipped
            if (invert) {
              // Increment count
              // of operations
              operations++;

              // Insert length of
              // the chosen prefix
              ops.push(i + 1);

              // Update invert to false
              invert = false;
            }
          }
        }

        // Print the number of
        // operations required
        document.write(operations + "<br>");

        // Print the chosen prefix
        // length in each operation
        if (operations !== 0) {
          for (const x of ops) {
            document.write(x + " ");
          }
        }
      }

      // Driver Code
      // Given binary Strings
      var A = "001",
        B = "000";
      var N = A.length;

      findOperations(A, B, N);
    </script>

2
3 2

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