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340. Longest Substring with At Most K Distinct Characters 👍

Approach 1: Sliding Window

  • Time: $O(n)$
  • Space: $O(128) = O(1)$
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class Solution {
 public:
  int lengthOfLongestSubstringKDistinct(string s, int k) {
    int ans = 0;
    int distinct = 0;
    vector<int> count(128);

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

    return ans;
  }
};
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class Solution {
  public int lengthOfLongestSubstringKDistinct(String s, int k) {
    int ans = 0;
    int distinct = 0;
    int[] count = new int[128];

    for (int l = 0, r = 0; r < s.length(); ++r) {
      if (++count[s.charAt(r)] == 1)
        ++distinct;
      while (distinct == k + 1)
        if (--count[s.charAt(l++)] == 0)
          --distinct;
      ans = Math.max(ans, r - l + 1);
    }

    return ans;
  }
}
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class Solution:
  def lengthOfLongestSubstringKDistinct(self, s: str, k: int) -> int:
    ans = 0
    distinct = 0
    count = collections.Counter()

    l = 0
    for r, c in enumerate(s):
      count[c] += 1
      if count[c] == 1:
        distinct += 1
      while distinct == k + 1:
        count[s[l]] -= 1
        if count[s[l]] == 0:
          distinct -= 1
        l += 1
      ans = max(ans, r - l + 1)

    return ans

Approach 2: Ordered Map

  • Time: $O(n\log k)$
  • Space: $O(n)$
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class Solution {
 public:
  int lengthOfLongestSubstringKDistinct(string s, int k) {
    int ans = 0;
    map<int, char> lastSeen;          // {last index: char}
    unordered_map<char, int> window;  // {char: index}

    for (int l = 0, r = 0; r < s.length(); ++r) {
      const int inChar = s[r];
      if (window.count(inChar))
        lastSeen.erase(window[inChar]);
      lastSeen[r] = inChar;
      window[inChar] = r;
      if (window.size() > k) {
        const auto [lastIndex, outChar] = *begin(lastSeen);
        lastSeen.erase(begin(lastSeen));
        window.erase(outChar);
        l = lastIndex + 1;
      }
      ans = max(ans, r - l + 1);
    }

    return ans;
  }
};