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76. Minimum Window Substring 👍

  • Time: $O(|\texttt{s}| + |\texttt{t}|)$
  • Space: $O(128) = O(1)$
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class Solution {
 public:
  string minWindow(string s, string t) {
    vector<int> count(128);
    int required = t.length();
    int bestLeft = -1;
    int minLength = s.length() + 1;

    for (const char c : t)
      ++count[c];

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

    return bestLeft == -1 ? "" : s.substr(bestLeft, minLength);
  }
};

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class Solution {
  public String minWindow(String s, String t) {
    int[] count = new int[128];
    int required = t.length();
    int bestLeft = -1;
    int minLength = s.length() + 1;

    for (final char c : t.toCharArray())
      ++count[c];

    for (int l = 0, r = 0; r < s.length(); ++r) {
      if (--count[s.charAt(r)] >= 0)
        --required;
      while (required == 0) {
        if (r - l + 1 < minLength) {
          bestLeft = l;
          minLength = r - l + 1;
        }
        if (++count[s.charAt(l++)] > 0)
          ++required;
      }
    }

    return bestLeft == -1 ? "" : s.substring(bestLeft, bestLeft + minLength);
  }
}

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class Solution:
  def minWindow(self, s: str, t: str) -> str:
    count = collections.Counter(t)
    required = len(t)
    bestLeft = -1
    minLength = len(s) + 1

    l = 0
    for r, c in enumerate(s):
      count[c] -= 1
      if count[c] >= 0:
        required -= 1
      while required == 0:
        if r - l + 1 < minLength:
          bestLeft = l
          minLength = r - l + 1
        count[s[l]] += 1
        if count[s[l]] > 0:
          required += 1
        l += 1

    return '' if bestLeft == -1 else s[bestLeft: bestLeft + minLength]