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1297. Maximum Number of Occurrences of a Substring

  • Time: $O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  int maxFreq(string s, int maxLetters, int minSize, int maxSize) {
    // Greedily consider strings with `minSize`, so ignore `maxSize`.
    int ans = 0;
    int letters = 0;
    vector<int> count(26);
    unordered_map<string, int> substringCount;

    for (int l = 0, r = 0; r < s.length(); ++r) {
      if (++count[s[r] - 'a'] == 1)
        ++letters;
      while (letters > maxLetters || r - l + 1 > minSize)
        if (--count[s[l++] - 'a'] == 0)
          --letters;
      if (r - l + 1 == minSize)
        ans = max(ans, ++substringCount[s.substr(l, minSize)]);
    }

    return ans;
  }
};

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class Solution {
  public int maxFreq(String s, int maxLetters, int minSize, int maxSize) {
    // Greedily consider strings with `minSize`, so ignore `maxSize`.
    int ans = 0;
    int letters = 0;
    int[] count = new int[26];
    Map<String, Integer> substringCount = new HashMap<>();

    for (int l = 0, r = 0; r < s.length(); ++r) {
      if (++count[s.charAt(r) - 'a'] == 1)
        ++letters;
      while (letters > maxLetters || r - l + 1 > minSize)
        if (--count[s.charAt(l++) - 'a'] == 0)
          --letters;
      if (r - l + 1 == minSize)
        ans = Math.max(ans, substringCount.merge(s.substring(l, l + minSize), 1, Integer::sum));
    }

    return ans;
  }
}

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class Solution:
  def maxFreq(self, s: str, maxLetters: int, minSize: int, maxSize: int) -> int:
    # Greedily consider strings with `minSize`, so ignore `maxSize`.
    ans = 0
    letters = 0
    count = collections.Counter()
    substringCount = collections.Counter()

    l = 0
    for r, c in enumerate(s):
      count[c] += 1
      if count[c] == 1:
        letters += 1
      while letters > maxLetters or r - l + 1 > minSize:
        count[s[l]] -= 1
        if count[s[l]] == 0:
          letters -= 1
        l += 1
      if r - l + 1 == minSize:
        sub = s[l:l + minSize]
        substringCount[sub] += 1
        ans = max(ans, substringCount[sub])

    return ans