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3258. Count Substrings That Satisfy K-Constraint I 👍

  • Time: $O(n)$
  • Space: $O(1)$
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class Solution {
 public:
  int countKConstraintSubstrings(string s, int k) {
    int ans = 0;
    vector<int> count(2);

    for (int l = 0, r = 0; r < s.length(); ++r) {
      ++count[s[r] - '0'];
      while (count[0] > k && count[1] > k)
        --count[s[l++] - '0'];
      ans += r - l + 1;
    }

    return ans;
  }
};

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class Solution {
  public int countKConstraintSubstrings(String s, int k) {
    int ans = 0;
    int[] count = new int[2];

    int l = 0;
    for (int r = 0; r < s.length(); ++r) {
      ++count[s.charAt(r) - '0'];
      while (count[0] > k && count[1] > k)
        --count[s.charAt(l++) - '0'];
      ans += r - l + 1;
    }

    return ans;
  }
}

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class Solution:
  def countKConstraintSubstrings(self, s: str, k: int) -> int:
    ans = 0
    count = [0, 0]

    l = 0
    for r, c in enumerate(s):
      count[int(c)] += 1
      while min(count) > k:
        count[int(s[l])] -= 1
        l += 1
      ans += r - l + 1

    return ans