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76. Minimum Window Substring

Difficulty Topics

Description

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.

A 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?

Solution

minimum-window-substring.py
class Solution:
    def minWindow(self, s: str, t: str) -> str:
        sn, tn = len(s), len(t)
        targetCounter = Counter(t)
        targetLength = len(targetCounter)
        res = (inf, -1)
        i = 0

        for j, x in enumerate(s):
            if x in targetCounter:
                targetCounter[x] -= 1
                if targetCounter[x] == 0:
                    targetLength -= 1

            while targetLength == 0:
                currSize = j - i + 1
                if currSize < res[0]:
                    res = (currSize, i)

                if s[i] in targetCounter:
                    targetCounter[s[i]] += 1
                    if targetCounter[s[i]] == 1:
                        targetLength += 1

                i += 1

        return "" if res[0] == inf else s[res[1]: res[0] + res[1]]