给定二进制字符串中仅包含 1 的 K 长度子数组的计数 |设置 2

给定二进制字符串str ，任务是找到仅包含1的K个长度子数组的计数。

例子

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

方法：给定的问题也可以通过使用滑动窗口技术来解决。创建一个大小为K的窗口，初始计数为1秒，范围为 0到K-1 。然后从索引1到N-1遍历字符串并减去i-1的值并将i+K的值添加到当前计数。在这里，如果当前计数等于K ，则增加子数组的可能计数。

下面是上述方法的实现。

C++

// C++ program for the above approach
#include 
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

   // 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
 

输出

3

时间复杂度： O(N)，其中 N 是字符串的长度。

辅助空间： O(1)。