给定二进制字符串中仅包含 1 的 K 个长度子数组的计数

给定一个二进制字符串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 with 1 ones

Input: str = “11111001”, K=3
Output: 3

方法：这个任务可以通过跟踪连续的组大小来解决。一旦我们得到groupSize ，我们可以推断出长度为k的可能子数组的数量，并且所有1都是groupSize – k + 1 。

请按照以下步骤解决问题：

• 从头开始迭代二进制字符串
• 增加计数，如果遇到1 ，并且在0出现的点。
• 存储当前计数得到连续1的groupSize，重新初始化计数为0。
• 使用关系groupSize – k + 1在此groupSize中添加大小为 k 的可能子数组的计数
• 返回计数的最终总和。

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)
{
    // Add dummy character at last to handle
    // edge cases, where string ends with '1'
    s += '0';
    int n = s.length();
 
    int cnt = 0, ans = 0;
    for (int i = 0; i < n; i++) {
        if (s[i] == '1')
            cnt++;
        else {
            if (cnt >= k) {
                ans += (cnt - k + 1);
            }
            cnt = 0;
        }
    }
    return ans;
}
 
// Driver code
int main()
{
    string str = "0101000";
    int K = 1;
    cout << get(str, K) << endl;
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find the count of all possible
// k length subarrays
static int get(String s, int k)
{
   
    // Add dummy character at last to handle
    // edge cases, where String ends with '1'
    s += '0';
    int n = s.length();
 
    int cnt = 0, ans = 0;
    for (int i = 0; i < n; i++) {
        if (s.charAt(i) == '1')
            cnt++;
        else {
            if (cnt >= k) {
                ans += (cnt - k + 1);
            }
            cnt = 0;
        }
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    String str = "0101000";
    int K = 1;
    System.out.print(get(str, K) +"\n");
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python code for the above approach
 
# Function to find the count of all possible
# k length subarrays
def get(s, k):
 
    # Add dummy character at last to handle
    # edge cases, where string ends with '1'
    s += '0'
    n = len(s)
 
    cnt = 0
    ans = 0
    for i in range(n):
        if (s[i] == '1'):
            cnt += 1
        else:
            if (cnt >= k):
                ans += (cnt - k + 1)
            cnt = 0
    return ans
 
# Driver code
str = "0101000"
K = 1
print(get(str, K))
 
# This code is contributed by Saurabh Jaiswal

C#

// C# program for the above approach
using System;
class GFG
{
   
    // Function to find the count of all possible
    // k length subarrays
    static int get(string s, int k)
    {
       
        // Add dummy character at last to handle
        // edge cases, where string ends with '1'
        s += '0';
        int n = s.Length;
 
        int cnt = 0, ans = 0;
        for (int i = 0; i < n; i++) {
            if (s[i] == '1')
                cnt++;
            else {
                if (cnt >= k) {
                    ans += (cnt - k + 1);
                }
                cnt = 0;
            }
        }
        return ans;
    }
 
    // Driver code
    public static void Main()
    {
        string str = "0101000";
        int K = 1;
        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)
    {
     
      // Add dummy character at last to handle
      // edge cases, where string ends with '1'
      s += '0';
      let n = s.length;
 
      let cnt = 0, ans = 0;
      for (let i = 0; i < n; i++) {
        if (s[i] == '1')
          cnt++;
        else {
          if (cnt >= k) {
            ans += (cnt - k + 1);
          }
          cnt = 0;
        }
      }
      return ans;
    }
 
    // Driver code
    let str = "0101000";
    let K = 1;
    document.write(get(str, K) + '<br>');
 
  // This code is contributed by Potta Lokesh
  

输出

2

时间复杂度： O(N)
辅助空间： O(1)