Minimum prefixes required to be flipped to convert a Binary String to another
Last Updated : 28 May, 2021
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 ith 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++ // 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> Time Complexity: O(N)
Auxiliary Space: O(N)
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