A general method for substring p

Given a binary string of length N and an integer K, we need to find out how many substrings of this string are exist which contains exactly K ones.
Examples:
Input : s = “10010”
K = 1
Output : 9
The 9 substrings containing one 1 are,
“1”, “10”, “100”, “001”, “01”, “1”,
“10”, “0010” and “010”

static int countOfSubstringWithKOnes( 
                            String s, int K) 
    { 
        int N = s.length(); 
        int res = 0; 
        int countOfOne = 0; 
        int []freq = new int[N+1]; 

        freq[0] = 1; 

        for (int i = 0; i < N; i++) { 
            countOfOne += (s.charAt(i) - '0'); 
            if (countOfOne >= K) { 
                res += freq[countOfOne - K]; 
            } 
            freq[countOfOne]++; 
        }   
        return res; 
    } 

注意：求子串的时候，尤其是涉及长度，子串里面元素个数之类的问题，可以用类似于动态规划的方法，用当前遍历的值减去之前的值，来找子串，实现复杂度O(n)。
相似的问题：https://www.geeksforgeeks.org/longest-substring-with-count-of-1s-more-than-0s/