Minimum Window Substring LeetCode Solution | Easy Approach

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Given two strings s and t of lengths m and n respectively, return the minimum window substring of s such that every character in t (including duplicates) is included in the window. If there is no such substring, return the empty string "".

The testcases will be generated such that the answer is unique.

substring is a contiguous sequence of characters within the string.

Example 1:

Input: s = "ADOBECODEBANC", t = "ABC"
Output: "BANC"
Explanation: The minimum window substring "BANC" includes 'A', 'B', and 'C' from string t.

Example 2:

Input: s = "a", t = "a"
Output: "a"
Explanation: The entire string s is the minimum window.

Example 3:

Input: s = "a", t = "aa"
Output: ""
Explanation: Both 'a's from t must be included in the window.
Since the largest window of s only has one 'a', return empty string.

Constraints:

  • m == s.length
  • n == t.length
  • 1 <= m, n <= 105
  • s and t consist of uppercase and lowercase English letters.

Follow up: Could you find an algorithm that runs in O(m + n) time?

Minimum Window SubstringSolutions

Time:  O(∣s∣+∣t∣)
Space: O(128)=O(1)

C++

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);
  }
};

Java

 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);
  }
}

Python

 class Solution:
  def minWindow(self, s: str, t: str) -> str:
    count = 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]

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