Minimum Window Substring

76. Minimum Window Substring

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.

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?

Analysis

Solution

class Solution:
    def minWindow(self, s: str, t: str) -> str:
        need = Counter()
        for x in t:
            need[x] += 1
        l = 0
        r = 0
        valid = 0
        start = 0
        n = len(s)
        length = inf
        w = Counter()
        while r < n:
            c = s[r]
            r += 1
            if need[c]:
                w[c] += 1
                if w[c] == need[c]: # only add the count if the window alreay contain enough number of a char
                    valid += 1
            while valid == len(need):
                if r - l < length:
                    start = l
                    length = r - l
                d = s[l]
                l += 1
                if need[d]:
                    if w[d] == need[d]: # only decrement the count if the left-most char's count decrease below the required
                        valid -= 1
                    w[d] -= 1
        if length == inf:
            return ''
        return s[start:start+length]