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Count of K length subarrays containing only 1s in given Binary String | Set 2

Last Updated : 23 Jul, 2025
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Given binary string str, the task is to find the count of K length subarrays containing only 1s.

Examples

Input: str = "0101000", K=1
Output: 2
Explanation: 0101000 -> There are 2 subarrays of length 1 containing only 1s.

Input: str = "11111001", K=3
Output: 3

 

Approach: The given problem can also be solved by using the Sliding Window technique. Create a window of size K initially with the count of 1s from range 0 to K-1. Then traverse the string from index 1 to N-1 and subtract the value of i-1 and add the value of i+K to the current count. Here if the current count is equal to K, increment the possible count of subarrays. 

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 count of all possible
// k length subarrays
int get(string s, int k)
{
    int n = s.length();

    int cntOf1s = 0;

    for (int i = 0; i < k; i++)
        if (s[i] == '1')
            cntOf1s++;

    int ans = cntOf1s == k ? 1 : 0;

    for (int i = 1; i < n; i++) {
        cntOf1s = cntOf1s - (s[i - 1] - '0')
                  + (s[i + k - 1] - '0');
        if (cntOf1s == k)
            ans++;
    }
    return ans;
}

// Driver code
int main()
{
    string str = "0110101110";
    int K = 2;
    cout << get(str, K) << endl;
    return 0;
}
Java 
// Java code to implement above approach
import java.util.*;
public class GFG {

  // Function to find the count of all possible
  // k length subarrays
  static int get(String s, int k)
  {
    int n = s.length();

    int cntOf1s = 0;

    for (int i = 0; i < k; i++) {
      if (s.charAt(i) == '1') {
        cntOf1s++;
      }
    }

    int ans = cntOf1s == k ? 1 : 0;

    for (int i = 1; i < n; i++) {
      if(i + k - 1 < n) {
        cntOf1s = cntOf1s - (s.charAt(i - 1) - '0')
          + (s.charAt(i + k - 1) - '0');
      }
      if (cntOf1s == k)
        ans++;
    }
    return ans;
  }

  // Driver code
  public static void main(String args[])
  {
    String str = "0110101110";
    int K = 2;
    System.out.println(get(str, K));

  }
}

// This code is contributed by Samim Hossain Mondal.
Python3 
# Python code to implement above approach

# Function to find the count of all possible
# k length subarrays
def get(s, k):
    n = len(s);

    cntOf1s = 0;

    for i in range(0,k):
        if (s[i] == '1'):
            cntOf1s += 1;

    ans = i if (cntOf1s == k) else 0;

    for i in range(1, n):
        if (i + k - 1 < n):
            cntOf1s = cntOf1s - (ord(s[i - 1]) - ord('0')) + (ord(s[i + k - 1]) - ord('0'));

        if (cntOf1s == k):
            ans += 1;

    return ans;

# Driver code
if __name__ == '__main__':
    str = "0110101110";
    K = 2;
    print(get(str, K));

# This code is contributed by 29AjayKumar
C# 
// C# code to implement above approach
using System;

public class GFG {

  // Function to find the count of all possible
  // k length subarrays
  static int get(string s, int k)
  {
    int n = s.Length;

    int cntOf1s = 0;

    for (int i = 0; i < k; i++) {
      if (s[i] == '1') {
        cntOf1s++;
      }
    }

    int ans = cntOf1s == k ? 1 : 0;

    for (int i = 1; i < n; i++) {
      if (i + k - 1 < n) {
        cntOf1s = cntOf1s - (s[i - 1] - '0')
          + (s[i + k - 1] - '0');
      }
      if (cntOf1s == k)
        ans++;
    }
    return ans;
  }

  // Driver code
  public static void Main()
  {
    string str = "0110101110";
    int K = 2;
    Console.WriteLine(get(str, K));
  }
}

// This code is contributed by ukasp.
JavaScript 
 <script>
    // JavaScript code for the above approach

    // Function to find the count of all possible
    // k length subarrays
    function get(s, k)
    {
      let n = s.length;
      let cntOf1s = 0;

      for (let i = 0; i < k; i++)
        if (s[i] == '1')
          cntOf1s++;

      let ans = cntOf1s == k ? 1 : 0;

      for (let i = 1; i < n; i++) 
      {
        cntOf1s = cntOf1s - (s[i - 1] - '0')
          + (s[i + k - 1] - '0');
        if (cntOf1s == k)
          ans++;
      }
      return ans;
    }

    // Driver code
    let str = "0110101110";
    let K = 2;
    document.write(get(str, K) + '<br>')

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

 
 


Output
3


 Time Complexity: O(N), Where N is the length of the string. 

Auxiliary Space: O(1). 


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