Minimize operations to make given binary string as all 1s by repeatedly converting K consecutive characters to 1

Given a binary string str of N characters an integer K, the task is to find the minimum moves required to convert all characters of the string to 1’s where at each move, K consecutive characters can be converted to 1.

Example:

Input: str=”0010″, K=3
Output: 2
Explanation:
Move 1: Select the substring from 0 to 2 and replace it with all 1.
Move 2: Select the substring from 1 to 3 and replace it with all 1.

Input: str=”0000010″, K=1
Output: 6

Approach: This problem can be solved using a greedy approach. To solve this problem, follow the below steps:

1. Create a variable cnt, to store the minimum number of moves required. Initialize it with 0.
2. Traverse the string using a variable i and str[i] = ‘0’, then update i to i + K and increase cnt by 1, because no matter what these characters are, a move is always required to convert the character str[i] to ‘1’.
3. After the loop ends, return cnt which will be the required answer.

Below is the implementation of the above approach:

C++

// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the minimum
// operations required to convert
// the binary string str to all 1s
int minMoves(string str, int K)
{
    int N = str.size();

    // Variable to store number
    // of minimum moves required
    int cnt = 0;

    int i = 0;

    // Loop to traverse str
    while (i < N) {

        // If element is '0'
        if (str[i] == '0') {
            i += K;
            cnt += 1;
        }

        // If element is '1'
        else {
            i++;
        }
    }

    return cnt;
}

// Driver Code
int main()
{
    string str = "0010";
    int K = 3;
    cout << minMoves(str, K);
}

Java

// Java code for the above approach
class GFG {

    // Function to find the minimum
    // operations required to convert
    // the binary String str to all 1s
    static int minMoves(String str, int K) {
        int N = str.length();

        // Variable to store number
        // of minimum moves required
        int cnt = 0;

        int i = 0;

        // Loop to traverse str
        while (i < N) {

            // If element is '0'
            if (str.charAt(i) == '0') {
                i += K;
                cnt += 1;
            }

            // If element is '1'
            else {
                i++;
            }
        }

        return cnt;
    }

    // Driver Code
    public static void main(String args[]) {
        String str = "0010";
        int K = 3;
        System.out.println(minMoves(str, K));
    }
}

// This code is contributed by Saurabh Jaiswal

Python3

# Python code for the above approach

# Function to find the minimum
# operations required to convert
# the binary string str to all 1s
def minMoves(str, K):
    N = len(str)

    # Variable to store number
    # of minimum moves required
    cnt = 0

    i = 0

    # Loop to traverse str
    while (i < N):

        # If element is '0'
        if (str[i] == '0'):
            i += K
            cnt += 1
        # If element is '1'
        else:
            i += 1

    return cnt

# Driver Code
str = "0010"
K = 3
print(minMoves(str, K))

# This code is contributed by Saurabh Jaiswal

C#

// C# code for the above approach
using System;

class GFG
{

// Function to find the minimum
// operations required to convert
// the binary string str to all 1s
static int minMoves(string str, int K)
{
    int N = str.Length;

    // Variable to store number
    // of minimum moves required
    int cnt = 0;

    int i = 0;

    // Loop to traverse str
    while (i < N) {

        // If element is '0'
        if (str[i] == '0') {
            i += K;
            cnt += 1;
        }

        // If element is '1'
        else {
            i++;
        }
    }

    return cnt;
}

// Driver Code
public static void Main()
{
    string str = "0010";
    int K = 3;
    Console.Write(minMoves(str, K));
}
}

// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
       // JavaScript code for the above approach


       // Function to find the minimum
       // operations required to convert
       // the binary string str to all 1s
       function minMoves(str, K) {
           let N = str.length;

           // Variable to store number
           // of minimum moves required
           let cnt = 0;

           let i = 0;

           // Loop to traverse str
           while (i < N) {

               // If element is '0'
               if (str[i] == '0') {
                   i += K;
                   cnt += 1;
               }

               // If element is '1'
               else {
                   i++;
               }
           }

           return cnt;
       }

       // Driver Code

       let str = "0010";
       let K = 3;
       document.write(minMoves(str, K));

 // This code is contributed by Potta Lokesh
   </script>

Output

2

Time Complexity: O(N)
Auxiliary Space: O(1)