Skip to content
Related Articles

Related Articles

Improve Article

Minimize flips to make binary string as all 1s by flipping characters in substring of size K repeatedly

  • Last Updated : 29 Sep, 2021
Geek Week

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:

  1. Consider the substring S[0, 2] which is of size K(= 3) and flipping those characters modifies the string to “11110110”.
  2. Consider the substring S[4, 6] which is of size K(= 3) and flipping those characters modifies the string to “11111000”.
  3. 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 K
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[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 Code
int main()
{
 
    string S = "00010110";
    int K = 3;
    int N = S.length();
    minOperation(S, K, N);
 
    return 0;
}

Java




// Java program for the above approach
public class GFG
{
 
// Function to find the minimum number
// of operations required to convert all
// the characters to 1 by flipping the
// substrings of size K
static 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 Code
public 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 K
def 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 Code
S = "00010110"
K = 3
N = len(S)
minOperation(S, K, N)
 
# This code is contributed by gfgking

C#




// C# program for the above approach
using 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 K
static 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 Code
public 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 K
function 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)

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :