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# 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 substrings of a binary string containing K ones

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

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