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Minimum number of flips or swaps of adjacent characters required to make two strings equal

  • Last Updated : 23 Aug, 2021

Given two binary strings A and B of length N, the task is to count the minimum number of operations required to make the two given strings equal by either swapping adjacent characters or flipping any character of the string A.

Examples:

Input: A = “100”, B = “001”
Output: 2
Explanation: Flipping characters A[0](= ‘1’) and A[2](= ‘0’) modifies the string A to “001”, which is equal to the string B.

Input: A = “0101”, B = “0011”
Output: 1
Explanation: Swapping the characters A[2](= ‘0’) and A[3](= ‘1’) modifies the string A to “0011”, which is equal to string B.

Approach: The problem can be solved using Dynamic Programming as it has Overlapping Subproblems and Optimal Substructure
Follow the steps below to solve the problem:



  • Initialize an array, say dp[] of size N+1 as {0}, where dp[i] stores the minimum number of operations required up to index i, to make the prefix of Ai equal to the prefix Bi.
  • Iterate over the range [1, N] using a variable, say i, and performing the following operations:
    • If A[i – 1] equals B[i – 1], then update dp[i] to dp[i – 1].
    • Otherwise, update dp[i] to dp[i – 1] + 1.
    • If swapping is possible, i.e. i > 1 and A[i – 2] is equal to B[i – 1] and A[i – 1] is equal to B[i – 2], then update dp[i] to min(dp[i], dp[i – 2] + 1).
  • After completing the above steps, print the minimum number of operations obtained, i.e. the value dp[N].

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the minimum
// number of operations required
// to make strings A and B equal
int countMinSteps(string A, string B, int N)
{
 
    // Stores all dp-states
    vector<int> dp(N + 1, 0);
 
    // Iterate over the range [1, N]
    for (int i = 1; i <= N; i++) {
 
        // If A[i - 1] equals to B[i - 1]
        if (A[i - 1] == B[i - 1]) {
 
            // Assign Dp[i - 1] to Dp[i]
            dp[i] = dp[i - 1];
        }
 
        // Otherwise
        else {
 
            // Update dp[i]
            dp[i] = dp[i - 1] + 1;
        }
 
        // If swapping is possible
        if (i >= 2 && A[i - 2] == B[i - 1]
            && A[i - 1] == B[i - 2]) {
 
            // Update dp[i]
            dp[i] = min(dp[i], dp[i - 2] + 1);
        }
    }
 
    // Return the minimum
    // number of steps required
    return dp[N];
}
 
// Driver Code
int main()
{
    // Given Input
    string A = "0101";
    string B = "0011";
    int N = A.length();
 
    // Function Call
    cout << countMinSteps(A, B, N);
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
 
class GFG{
     
// Function to count the minimum
// number of operations required
// to make strings A and B equal
static int countMinSteps(String A, String B, int N)
{
     
    // Stores all dp-states
    int[] dp = new int[N + 1];
    for(int i = 1; i <= N; i++)
    {
         
        // Update the value of A[i]
        dp[i] = 0;
    }
 
    // Iterate over the range [1, N]
    for(int i = 1; i <= N; i++)
    {
         
        // If A[i - 1] equals to B[i - 1]
        if (A.charAt(i - 1) == B.charAt(i - 1))
        {
             
            // Assign Dp[i - 1] to Dp[i]
            dp[i] = dp[i - 1];
        }
 
        // Otherwise
        else
        {
             
            // Update dp[i]
            dp[i] = dp[i - 1] + 1;
        }
 
        // If swapping is possible
        if (i >= 2 && A.charAt(i - 2) == B.charAt(i - 1) &&
                      A.charAt(i - 1) == B.charAt(i - 2))
        {
             
            // Update dp[i]
            dp[i] = Math.min(dp[i], dp[i - 2] + 1);
        }
    }
 
    // Return the minimum
    // number of steps required
    return dp[N];
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given Input
    String A = "0101";
    String B = "0011";
    int N = A.length();
     
    // Function Call
    System.out.println(countMinSteps(A, B, N));
}
}
 
// This code is contributed by sanjoy_62

Python3




# Python3 program for the above approach
 
# Function to count the minimum
# number of operations required
# to make strings A and B equal
def countMinSteps(A, B, N) :
 
    # Stores all dp-states
    dp = [0] * (N + 1)
 
    # Iterate rate over the range [1, N]
    for i in range(1, N+1) :
 
        # If A[i - 1] equals to B[i - 1]
        if (A[i - 1] == B[i - 1]) :
 
            # Assign Dp[i - 1] to Dp[i]
            dp[i] = dp[i - 1]
         
 
        # Otherwise
        else :
 
            # Update dp[i]
            dp[i] = dp[i - 1] + 1
         
 
        # If swapping is possible
        if (i >= 2 and A[i - 2] == B[i - 1]
            and A[i - 1] == B[i - 2]) :
 
            # Update dp[i]
            dp[i] = min(dp[i], dp[i - 2] + 1)
         
    # Return the minimum
    # number of steps required
    return dp[N]
 
 
# Driver Code
 
# Given Input
A = "0101"
B = "0011"
N = len(A)
 
# Function Call
print(countMinSteps(A, B, N))
 
# This code is contributed by splevel62.

C#




// C# program for the above approach
using System;
 
class GFG{
 
// Function to count the minimum
// number of operations required
// to make strings A and B equal
static int countMinSteps(string A, string B, int N)
{
     
    // Stores all dp-states
    int[] dp = new int[N + 1];
    for(int i = 1; i <= N; i++)
    {
         
        // Update the value of A[i]
        dp[i] = 0;
    }
 
    // Iterate over the range [1, N]
    for(int i = 1; i <= N; i++)
    {
         
        // If A[i - 1] equals to B[i - 1]
        if (A[i - 1] == B[i - 1])
        {
             
            // Assign Dp[i - 1] to Dp[i]
            dp[i] = dp[i - 1];
        }
 
        // Otherwise
        else
        {
             
            // Update dp[i]
            dp[i] = dp[i - 1] + 1;
        }
 
        // If swapping is possible
        if (i >= 2 && A[i - 2] == B[i - 1] &&
                      A[i - 1] == B[i - 2])
        {
             
            // Update dp[i]
            dp[i] = Math.Min(dp[i], dp[i - 2] + 1);
        }
    }
 
    // Return the minimum
    // number of steps required
    return dp[N];
}
 
// Driver code
public static void Main(String []args)
{
     
    // Given Input
    string A = "0101";
    string B = "0011";
    int N = A.Length;
 
    // Function Call
    Console.Write(countMinSteps(A, B, N));
}
}
 
// This code is contributed by code_hunt

Javascript




<script>
       // JavaScript Program for the above approach
 
 
       // Function to count the minimum
       // number of operations required
       // to make strings A and B equal
       function countMinSteps(A, B, N) {
 
           // Stores all dp-states
           let dp = new Array(N + 1).fill(0);
 
           // Iterate over the range [1, N]
           for (let i = 1; i <= N; i++) {
 
               // If A[i - 1] equals to B[i - 1]
               if (A[i - 1] == B[i - 1]) {
 
                   // Assign Dp[i - 1] to Dp[i]
                   dp[i] = dp[i - 1];
               }
 
               // Otherwise
               else {
 
                   // Update dp[i]
                   dp[i] = dp[i - 1] + 1;
               }
 
               // If swapping is possible
               if (i >= 2 && A[i - 2] == B[i - 1]
                   && A[i - 1] == B[i - 2]) {
 
                   // Update dp[i]
                   dp[i] = Math.min(dp[i], dp[i - 2] + 1);
               }
           }
 
           // Return the minimum
           // number of steps required
           return dp[N];
       }
 
       // Driver Code
 
       // Given Input
       let A = "0101";
       let B = "0011";
       let N = A.length;
 
       // Function Call
       document.write(countMinSteps(A, B, N));
 
 
   // This code is contributed by Potta Lokesh
   </script>
Output: 
1

 

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

 

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