Given a binary string S and an integer K, the task is to find the minimum number of flips required to convert the given string into a concatenation of K-length equal sub-strings. It is given that the given string can be split into K-length substrings.
Examples:
Input: S = “101100101”, K = 3
Output: 1
Explanation:
Flip the ‘0’ from index 5 to ‘1’.
The resultant string is S = “101101101”.
It is the concatenation of substring “101”.
Hence, the minimum number of flips required is 1.Input: S = “10110111”, K = 4
Output: 2
Explanation:
Flip the ‘0’ and ‘1’ at indexes 4 and 5 respectively.
The resultant string is S = “10111011”.
It is the concatenation of the substring “1011”.
Hence, the minimum number of flips required is 2.
Approach:
The problem can be solved using Greedy Approach.
Follow the steps below:
- Iterate the given string with increments of K indices from each index and keep a count of the 0s and 1s.
- The character which occurs the minimum number of times must be flipped and keep incrementing that count.
- Perform the above steps for all the indices from 0 to K-1 to obtain the minimum number of flips required.
Below is the implementation of the above approach:
C++
// C++ Program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function that returns the minimum// number of flips to convert// the s into a concatenation// of K-length sub-stringint minOperations(string S, int K){ // Stores the result int ans = 0; // Iterate through string index for (int i = 0; i < K; i++) { // Stores count of 0s & 1s int zero = 0, one = 0; // Iterate making K jumps for (int j = i; j < S.size(); j += K) { // Count 0's if (S[j] == '0') zero++; // Count 1's else one++; } // Add minimum flips // for index i ans += min(zero, one); } // Return minimum number // of flips return ans;}// Driver Codeint main(){ string S = "110100101"; int K = 3; cout << minOperations(S, K); return 0;} |
Java
// Java program to implement // the above approach import java.io.*;class GFG{// Function that returns the minimum// number of flips to convert// the s into a concatenation// of K-length sub-stringpublic static int minOperations(String S, int K){ // Stores the result int ans = 0; // Iterate through string index for(int i = 0; i < K; i++) { // Stores count of 0s & 1s int zero = 0, one = 0; // Iterate making K jumps for(int j = i; j < S.length(); j += K) { // Count 0's if (S.charAt(j) == '0') zero++; // Count 1's else one++; } // Add minimum flips // for index i ans += Math.min(zero, one); } // Return minimum number // of flips return ans;}// Driver Codepublic static void main(String args[]){ String S = "110100101"; int K = 3; System.out.println(minOperations(S, K));}}// This code is contributed by grand_master |
Python3
# Python3 program to implement# the above approach# Function that returns the minimum# number of flips to convert the s# into a concatenation of K-length# sub-stringdef minOperations(S, K): # Stores the result ans = 0 # Iterate through string index for i in range(K): # Stores count of 0s & 1s zero, one = 0, 0 # Iterate making K jumps for j in range(i, len(S), K): # Count 0's if(S[j] == '0'): zero += 1 # Count 1's else: one += 1 # Add minimum flips # for index i ans += min(zero, one) # Return minimum number # of flips return ans# Driver codeif __name__ == '__main__': s = "110100101" K = 3 print(minOperations(s, K))# This code is contributed by Shivam Singh |
C#
// C# program to implement // the above approach using System;class GFG{// Function that returns the minimum// number of flips to convert// the s into a concatenation// of K-length sub-stringpublic static int minOperations(String S, int K){ // Stores the result int ans = 0; // Iterate through string index for(int i = 0; i < K; i++) { // Stores count of 0s & 1s int zero = 0, one = 0; // Iterate making K jumps for(int j = i; j < S.Length; j += K) { // Count 0's if (S[j] == '0') zero++; // Count 1's else one++; } // Add minimum flips // for index i ans += Math.Min(zero, one); } // Return minimum number // of flips return ans;}// Driver Codepublic static void Main(String []args){ String S = "110100101"; int K = 3; Console.WriteLine(minOperations(S, K));}}// This code is contributed by 29AjayKumar |
Javascript
<script> // JavaScript program to implement // the above approach // Function that returns the minimum // number of flips to convert // the s into a concatenation // of K-length sub-string function minOperations(S, K) { // Stores the result var ans = 0; // Iterate through string index for (var i = 0; i < K; i++) { // Stores count of 0s & 1s var zero = 0, one = 0; // Iterate making K jumps for (var j = i; j < S.length; j += K) { // Count 0's if (S[j] === "0") zero++; // Count 1's else one++; } // Add minimum flips // for index i ans += Math.min(zero, one); } // Return minimum number // of flips return ans; } // Driver Code var S = "110100101"; var K = 3; document.write(minOperations(S, K));</script> |
Output:
2
Time Complexity: O(N)
Auxiliary Space: O(1)
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