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3258. Count Substrings That Satisfy K-Constraint I

Description

You are given a binary string s and an integer k.

A binary string satisfies the k-constraint if either of the following conditions holds:

  • The number of 0's in the string is at most k.
  • The number of 1's in the string is at most k.

Return an integer denoting the number of substrings of s that satisfy the k-constraint.

 

Example 1:

Input: s = "10101", k = 1

Output: 12

Explanation:

Every substring of s except the substrings "1010", "10101", and "0101" satisfies the k-constraint.

Example 2:

Input: s = "1010101", k = 2

Output: 25

Explanation:

Every substring of s except the substrings with a length greater than 5 satisfies the k-constraint.

Example 3:

Input: s = "11111", k = 1

Output: 15

Explanation:

All substrings of s satisfy the k-constraint.

 

Constraints:

  • 1 <= s.length <= 50
  • 1 <= k <= s.length
  • s[i] is either '0' or '1'.

Solutions

Solution 1: Sliding Window

We use two variables $\textit{cnt0}$ and $\textit{cnt1}$ to record the number of $0$s and $1$s in the current window, respectively. We use $\textit{ans}$ to record the number of substrings that satisfy the $k$ constraint, and $l$ to record the left boundary of the window.

When we move the window to the right, if the number of $0$s and $1$s in the window both exceed $k$, we need to move the window to the left until the number of $0$s and $1$s in the window are both no greater than $k$. At this point, all substrings in the window satisfy the $k$ constraint, and the number of such substrings is $r - l + 1$, where $r$ is the right boundary of the window. We add this count to $\textit{ans}$.

Finally, we return $\textit{ans}$.

The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.

  • class Solution { public int countKConstraintSubstrings(String s, int k) { int cnt0 = 0, cnt1 = 0; int ans = 0, l = 0; for (int r = 0; r < s.length(); ++r) { int x = s.charAt(r) - '0'; cnt0 += x ^ 1; cnt1 += x; while (cnt0 > k && cnt1 > k) { int y = s.charAt(l++) - '0'; cnt0 -= y ^ 1; cnt1 -= y; } ans += r - l + 1; } return ans; } } 
  • class Solution { public: int countKConstraintSubstrings(string s, int k) { int cnt0 = 0, cnt1 = 0; int ans = 0, l = 0; for (int r = 0; r < s.length(); ++r) { int x = s[r] - '0'; cnt0 += x ^ 1; cnt1 += x; while (cnt0 > k && cnt1 > k) { int y = s[l++] - '0'; cnt0 -= y ^ 1; cnt1 -= y; } ans += r - l + 1; } return ans; } }; 
  • class Solution: def countKConstraintSubstrings(self, s: str, k: int) -> int: cnt0 = cnt1 = 0 ans = l = 0 for r, c in enumerate(s): cnt0 += int(c) ^ 1 cnt1 += int(c) while cnt0 > k and cnt1 > k: cnt0 -= int(s[l]) ^ 1 cnt1 -= int(s[l]) l += 1 ans += r - l + 1 return ans 
  • func countKConstraintSubstrings(s string, k int) (ans int) { cnt0, cnt1, l := 0, 0, 0 for r, c := range s { x := int(c - '0') cnt0 += x ^ 1 cnt1 += x for cnt0 > k && cnt1 > k { y := int(s[l] - '0') cnt0 -= y ^ 1 cnt1 -= y l++ } ans += r - l + 1 } return } 
  • function countKConstraintSubstrings(s: string, k: number): number { let [cnt0, cnt1, ans, l] = [0, 0, 0, 0]; for (let r = 0; r < s.length; ++r) { const x = s[r] === '1' ? 1 : 0; cnt0 += x ^ 1; cnt1 += x; while (cnt0 > k && cnt1 > k) { const y = s[l++] === '1' ? 1 : 0; cnt0 -= y ^ 1; cnt1 -= y; } ans += r - l + 1; } return ans; } 
  • impl Solution { pub fn count_k_constraint_substrings(s: String, k: i32) -> i32 { let mut cnt = [0; 2]; let mut l = 0; let mut ans = 0; let s = s.as_bytes(); for (r, &c) in s.iter().enumerate() { cnt[(c - b'0') as usize] += 1; while cnt[0] > k && cnt[1] > k { cnt[(s[l] - b'0') as usize] -= 1; l += 1; } ans += r - l + 1; } ans as i32 } }

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