Given a binary string S of size N and an integer K, the task is to find the minimum number of operations required to make all characters as 1s in the binary string by flipping characters in the substring of size K. If it is not possible to do so, then print “-1”.
Examples:
Input: S = “00010110 “, K = 3
Output: 3
Explanation:
Below are the operations performed:
- Consider the substring S[0, 2] which is of size K(= 3) and flipping those characters modifies the string to “11110110”.
- Consider the substring S[4, 6] which is of size K(= 3) and flipping those characters modifies the string to “11111000”.
- Consider the substring S[5, 7] which is of size K(= 3) and flipping those characters modifies the string to “11111111”.
After the above operations, all the 0s are converted to 1s, and the minimum number of operations required is 3.
Input: S = “01010”, K = 4
Output: -1
Approach: The given problem can be solved using the Greedy Approach. Follow the below steps to solve the problem:
- Initialize a variable, say minOperations as 0 that stores the count of the minimum number of operations required.
- Traverse the string S using the variable i and if the characters S[i] is ‘0’ and the value of (i + K) is at most K, then flip all the characters of the substring S[i, i + K] and increment the value of minOperations by 1.
- After the above operations, traverse the string S and if there exists any character ‘0’ then print “-1” and break out of the loop.
- If all the characters of the string S are 1s, then print the minOperations as the resultant minimum number of operations.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the minimum number// of operations required to convert all// the characters to 1 by flipping the// substrings of size Kvoid minOperation(string S, int K, int N){ // Stores the minimum number of // operations required int min = 0; int i; // Traverse the string S for (i = 0; i < N; i++) { // If the character is 0 if (S[i] == '0' && i + K <= N) { // Flip the substrings of // size K starting from i for (int j = i; j < i + K; j++) { if (S[j] == '1') S[j] = '0'; else S[j] = '1'; } // Increment the minimum count // of operations required min++; } } // After performing the operations // check if string S contains any 0s for (i = 0; i < N; i++) { if (S[i] == '0') break; } // If S contains only 1's if (i == N) cout << min; else cout << -1;}// Driver Codeint main(){ string S = "00010110"; int K = 3; int N = S.length(); minOperation(S, K, N); return 0;} |
Java
// Java program for the above approachpublic class GFG{// Function to find the minimum number// of operations required to convert all// the characters to 1 by flipping the// substrings of size Kstatic void minOperation(String S, int K, int N){ // Stores the minimum number of // operations required int min = 0; int i; // Traverse the string S for (i = 0; i < N; i++) { // If the character is 0 if (S.charAt(i) == '0' && i + K <= N) { // Flip the substrings of // size K starting from i for (int j = i; j < i + K; j++) { if (S.charAt(j) == '1') S = S.substring(0, j) + '0' + S.substring(j + 1); else S = S.substring(0, j) + '1' + S.substring(j + 1); } // Increment the minimum count // of operations required min++; } } // After performing the operations // check if string S contains any 0s for (i = 0; i < N; i++) { if (S.charAt(i) == '0') break; } // If S contains only 1's if (i == N) System.out.println(min); else System.out.println(-1);}// Driver Codepublic static void main(String []args){ String S = "00010110"; int K = 3; int N = S.length(); minOperation(S, K, N);}}// This code is contributed by AnkThon |
Python3
# Function to find the minimum number# of operations required to convert all# the characters to 1 by flipping the# substrings of size Kdef minOperation(St, K, N): S = list(St) # Stores the minimum number of # operations required min = 0 i = 0 # Traverse the string S for i in range(0, N): # If the character is 0 if (S[i] == '0' and i + K <= N): # Flip the substrings of # size K starting from i j = i while(j < i + K): if (S[j] == '1'): S[j] = '0' else: S[j] = '1' j += 1 # Increment the minimum count # of operations required min += 1 # After performing the operations # check if string S contains any 0s temp = 0 for i in range(0, N): temp += 1 if (S[i] == '0'): break # If S contains only 1's if (temp == N): print(min) else: print(-1)# Driver CodeS = "00010110"K = 3N = len(S)minOperation(S, K, N)# This code is contributed by gfgking |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{// Function to find the minimum number// of operations required to convert all// the characters to 1 by flipping the// substrings of size Kstatic void minOperation(string S, int K, int N){ // Stores the minimum number of // operations required int min = 0; int i; // Traverse the string S for (i = 0; i < N; i++) { // If the character is 0 if (S[i] == '0' && i + K <= N) { // Flip the substrings of // size K starting from i for (int j = i; j < i + K; j++) { if (S[j] == '1') S = S.Substring(0, j) + '0' + S.Substring(j + 1); else S = S.Substring(0, j) + '1' + S.Substring(j + 1); } // Increment the minimum count // of operations required min++; } } // After performing the operations // check if string S contains any 0s for (i = 0; i < N; i++) { if (S[i] == '0') break; } // If S contains only 1's if (i == N) Console.Write(min); else Console.Write(-1);}// Driver Codepublic static void Main(){ string S = "00010110"; int K = 3; int N = S.Length; minOperation(S, K, N);}}// This code is contributed by SURENDRA_GANGWAR. |
Javascript
<script>// Function to find the minimum number// of operations required to convert all// the characters to 1 by flipping the// substrings of size Kfunction minOperation(St, K, N){ S = St.split(''); // Stores the minimum number of // operations required let min = 0; let i; // Traverse the string S for (i = 0; i < N; i++) { // If the character is 0 if (S[i] == '0' && i + K <= N) { // Flip the substrings of // size K starting from i for (let j = i; j < i + K; j++) { if (S[j] == '1') S[j] = '0'; else S[j] = '1'; } // Increment the minimum count // of operations required min++; } } // After performing the operations // check if string S contains any 0s for (i = 0; i < N; i++) { if (S[i] == '0') break; } // If S contains only 1's if (i == N) document.write(min); else document.write(-1);}// Driver Code let S = "00010110"; let K = 3; let N = S.length; minOperation(S, K, N);// This code is contributed by sanjoy_62. </script> |
Output:
3
Time Complexity: O(N*K)
Auxiliary Space: O(1)
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