Skip to content

3298. Count Substrings That Can Be Rearranged to Contain a String II 👍

  • Time: $O(n)$
  • Space: $O(26) = O(1)$
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
class Solution {
 public:
  // Same as 3297. Count Substrings That Can Be Rearranged to Contain a String I
  long long validSubstringCount(string word1, string word2) {
    long ans = 0;
    int required = word2.length();
    vector<int> count(26);

    for (const char c : word2)
      ++count[c - 'a'];

    for (int l = 0, r = 0; r < word1.length(); ++r) {
      if (--count[word1[r] - 'a'] >= 0)
        --required;
      while (required == 0) {
        // Add valid substrings containing word1[l..r] to the answer. They are
        // word1[l..r], word1[l..r + 1], ..., word1[l..n - 1].
        ans += word1.length() - r;
        if (++count[word1[l++] - 'a'] > 0)
          ++required;
      }
    }

    return ans;
  }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
class Solution {
  // Same as 3297. Count Substrings That Can Be Rearranged to Contain a String I
  public long validSubstringCount(String word1, String word2) {
    long ans = 0;
    int required = word2.length();
    int[] count = new int[26];

    for (final char c : word2.toCharArray())
      ++count[c - 'a'];

    for (int l = 0, r = 0; r < word1.length(); ++r) {
      if (--count[word1.charAt(r) - 'a'] >= 0)
        --required;
      while (required == 0) {
        // Add valid substrings containing word1[l..r] to the answer. They are
        // word1[l..r], word1[l..r + 1], ..., word1[l..n - 1].
        ans += word1.length() - r;
        if (++count[word1.charAt(l++) - 'a'] > 0)
          ++required;
      }
    }

    return ans;
  }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
class Solution:
  # Same as 3297. Count Substrings That Can Be Rearranged to Contain a String I
  def validSubstringCount(self, word1: str, word2: str) -> int:
    ans = 0
    count = collections.Counter(word2)
    required = len(word2)

    l = 0
    for r, c in enumerate(word1):
      count[c] -= 1
      if count[c] >= 0:
        required -= 1
      while required == 0:
        # Add valid substrings containing word1[l..r] to the answer. They are
        # word1[l..r], word1[l..r + 1], ..., word1[l..n - 1].
        ans += len(word1) - r
        count[word1[l]] += 1
        if count[word1[l]] > 0:
          required += 1
        l += 1

    return ans