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

Last Updated : 29 Dec, 2021

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

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

Auxiliary Space: O(1).

Count of binary strings of given length consisting of at least one 1

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Practice Tags :

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

C++

Java

Python3

C#

Javascript

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