Question

340. Longest Substring with At Most K Distinct Characters

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Description

Given a string s and an integer k, return _the length of the longest __substring__ of_ s _that contains at most_ k _distinct characters_.

 

Example 1:

Input: s = "eceba", k = 2
Output: 3
Explanation: The substring is "ece" with length 3.

Example 2:

Input: s = "aa", k = 1
Output: 2
Explanation: The substring is "aa" with length 2.

 

Constraints:

• 1 <= s.length <= 5 * 104
• 0 <= k <= 50

Solutions

Solution 1

class Solution:
  def lengthOfLongestSubstringKDistinct(self, s: str, k: int) -> int:
    cnt = Counter()
    n = len(s)
    ans = j = 0
    for i, c in enumerate(s):
      cnt[c] += 1
      while len(cnt) > k:
        cnt[s[j]] -= 1
        if cnt[s[j]] == 0:
          cnt.pop(s[j])
        j += 1
      ans = max(ans, i - j + 1)
    return ans

class Solution {
  public int lengthOfLongestSubstringKDistinct(String s, int k) {
    Map<Character, Integer> cnt = new HashMap<>();
    int n = s.length();
    int ans = 0, j = 0;
    for (int i = 0; i < n; ++i) {
      char c = s.charAt(i);
      cnt.put(c, cnt.getOrDefault(c, 0) + 1);
      while (cnt.size() > k) {
        char t = s.charAt(j);
        cnt.put(t, cnt.getOrDefault(t, 0) - 1);
        if (cnt.get(t) == 0) {
          cnt.remove(t);
        }
        ++j;
      }
      ans = Math.max(ans, i - j + 1);
    }
    return ans;
  }
}

class Solution {
public:
  int lengthOfLongestSubstringKDistinct(string s, int k) {
    unordered_map<char, int> cnt;
    int n = s.size();
    int ans = 0, j = 0;
    for (int i = 0; i < n; ++i) {
      cnt[s[i]]++;
      while (cnt.size() > k) {
        if (--cnt[s[j]] == 0) {
          cnt.erase(s[j]);
        }
        ++j;
      }
      ans = max(ans, i - j + 1);
    }
    return ans;
  }
};

func lengthOfLongestSubstringKDistinct(s string, k int) (ans int) {
  cnt := map[byte]int{}
  j := 0
  for i := range s {
    cnt[s[i]]++
    for len(cnt) > k {
      cnt[s[j]]--
      if cnt[s[j]] == 0 {
        delete(cnt, s[j])
      }
      j++
    }
    ans = max(ans, i-j+1)
  }
  return
}