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3. Longest Substring Without Repeating Characters 👍

Approach 1: Sliding window

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

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

    return ans;
  }
};

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class Solution {
  public int lengthOfLongestSubstring(String s) {
    int ans = 0;
    int[] count = new int[128];

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

    return ans;
  }
}

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class Solution:
  def lengthOfLongestSubstring(self, s: str) -> int:
    ans = 0
    count = collections.Counter()

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

    return ans

Approach 2: Last seen

  • Time: $O(n)$
  • Space: $O(128) = O(1)$
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class Solution {
 public:
  int lengthOfLongestSubstring(string s) {
    int ans = 0;
    // The substring s[j + 1..i] has no repeating characters.
    int j = -1;
    // lastSeen[c] := the index of the last time c appeared
    vector<int> lastSeen(128, -1);

    for (int i = 0; i < s.length(); ++i) {
      // Update j to lastSeen[s[i]], so the window must start from j + 1.
      j = max(j, lastSeen[s[i]]);
      ans = max(ans, i - j);
      lastSeen[s[i]] = i;
    }

    return ans;
  }
};

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class Solution {
  public int lengthOfLongestSubstring(String s) {
    int ans = 0;
    // The substring s[j + 1..i] has no repeating characters.
    int j = -1;
    // lastSeen[c] := the index of the last time c appeared
    int[] lastSeen = new int[128];
    Arrays.fill(lastSeen, -1);

    for (int i = 0; i < s.length(); ++i) {
      // Update j to lastSeen[s.charAt(i)], so the window must start from j + 1.
      j = Math.max(j, lastSeen[s.charAt(i)]);
      ans = Math.max(ans, i - j);
      lastSeen[s.charAt(i)] = i;
    }

    return ans;
  }
}

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class Solution:
  def lengthOfLongestSubstring(self, s: str) -> int:
    ans = 0
    # The substring s[j + 1..i] has no repeating characters.
    j = -1
    # lastSeen[c] := the index of the last time c appeared
    lastSeen = {}

    for i, c in enumerate(s):
      # Update j to lastSeen[c], so the window must start from j + 1.
      j = max(j, lastSeen.get(c, -1))
      ans = max(ans, i - j)
      lastSeen[c] = i

    return ans