Given two binary strings A and B, the task is to find the minimum number of times a substring starting from the first character of A needs to be flipped, i.e. convert 1s to 0s and 0s to 1s, to convert A to B.
Examples:
Input: A = “0010”, B = “1011”
Output; 3
Explanation:
Step 1: Flip the entire string A. Therefore, A becomes “1101” .
Step 2: Flip the substring {A[0], A[2]}. Therefore, A becomes “0011” .
Step 3: Flip A[0]. Therefore, A becomes “1011” which is equal to B.
Therefore, the minimum number of operations required is 3.Input: A = “1010101”, B = “0011100”
Output: 5
Explanation:
Step 1: Flip the entiThehrefore, A becomes “0101010″
Step 2: Flip substring {A[0], A[5]}. Therefore, A becomes “1010100″
Step 3: Flip the substring {A[0], A[3]}. Therefore, A becomes “0101100″
Step 4: Flip the substring {A[0], A[2]}. Therefore, A becomes “1011100″
Step 5: Flip A[0]. Therefore, A becomes “0011100” which is equal to B.
Therefore, the minimum number of operations required is 5.
Approach: The idea is to initialize a variable that holds the last index at which character at A is different from the character at B. Then negate A from the 1st index to the last index. Repeat until both strings become equal. Follow the steps below to solve the problem:
- Initialize a variable last_index that holds the last index at which characters are different in A and B.
- Negate string A from the 1st index to last_index and increment the count of steps.
- Repeat the above steps until string A becomes equal to string B.
- Print the count of steps after both the strings are the same after performing the operations.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function that finds the minimum// number of operations required such// that string A and B are the samevoid findMinimumOperations(string a, string b){ // Stores the count of steps int step = 0; // Stores the last index whose // bits are not same int last_index; // Iterate until both string // are unequal while (a != b) { // Check till end of string to // find righmost unequals bit for (int i = 0; i < a.length(); i++) { // Update the last index if (a[i] != b[i]) { last_index = i; } } // Flipping characters up // to the last index for (int i = 0; i <= last_index; i++) { // Flip the bit a[i] = (a[i] == '0') ? '1' : '0'; } // Increasing steps by one step++; } // Print the count of steps cout << step;}// Driver Codeint main(){ // Given strings A and B string A = "101010", B = "110011"; // Function Call findMinimumOperations(A, B); return 0;} |
Java
// Java program for the // above approachimport java.util.*;class GFG{// Function that finds the minimum// number of operations required such// that String A and B are the samestatic void findMinimumOperations(char[] a, char[] b){ // Stores the count of steps int step = 0; // Stores the last index whose // bits are not same int last_index = 0; // Iterate until both String // are unequal while (!Arrays.equals(a, b)) { // Check till end of String to // find righmost unequals bit for (int i = 0; i < a.length; i++) { // Update the last index if (a[i] != b[i]) { last_index = i; } } // Flipping characters up // to the last index for (int i = 0; i <= last_index; i++) { // Flip the bit a[i] = (a[i] == '0') ? '1' : '0'; } // Increasing steps by one step++; } // Print the count of steps System.out.print(step);}// Driver Codepublic static void main(String[] args){ // Given Strings A and B String A = "101010", B = "110011"; // Function Call findMinimumOperations(A.toCharArray(), B.toCharArray());}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program for the above approach# Function that finds the minimum# number of operations required such# that string A and B are the samedef findMinimumOperations(a, b): # Stores the count of steps step = 0 # Stores the last index whose # bits are not same last_index = 0 # Iterate until both string # are unequal while (a != b): a = [i for i in a] # Check till end of string to # find righmost unequals bit for i in range(len(a)): # Update the last index if (a[i] != b[i]): last_index = i # Flipping characters up # to the last index for i in range(last_index + 1): # Flip the bit if (a[i] == '0'): a[i] = '1' else: a[i] = '0' a = "".join(a) # Increasing steps by one step += 1 # Print the count of steps print(step)# Driver Codeif __name__ == '__main__': # Given strings A and B A = "101010" B = "110011" # Function Call findMinimumOperations(A, B)# This code is contributed by mohit kumar 29 |
C#
// C# program for the // above approachusing System;class GFG{// Function that finds the minimum// number of operations required such// that string A and B are the samestatic void findMinimumOperations(string a, string b){ // Stores the count of steps int step = 0; // Stores the last index whose // bits are not same int last_index = 0; // Iterate until both string // are unequal while (a.Equals(b) == false) { // Check till end of string to // find righmost unequals bit for(int i = 0; i < a.Length; i++) { // Update the last index if (a[i] != b[i]) { last_index = i; } } // Flipping characters up // to the last index char[] ch = a.ToCharArray(); for(int i = 0; i <= last_index; i++) { // Flip the bit if (ch[i] == '0') { ch[i] = '1'; } else { ch[i] = '0'; } } a = new string(ch); // Increasing steps by one step++; } // Print the count of steps Console.WriteLine(step);}// Driver Codepublic static void Main(){ // Given strings A and B string A = "101010"; string B = "110011"; // Function Call findMinimumOperations(A,B);}}// This code is contributed by SURENDRA_GANGWAR |
4
Time Complexity: O(N2)
Auxiliary Space: O(1)
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