Problem

Given two strings s and t of lengths m and n respectively, return the minimum window

substring

of

1
s

such that every character in

1
t

(including duplicates) is included in the window

. If there is no such substring, return

the empty string

1
""

.

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

Example 1:

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

Example 2:

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

Example 3:

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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?

Solve

目前沒有優化
runtime和別人有些差距

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class Solution:
    def minWindow(self, s: str, t: str) -> str:
        if s == t :
            return s

        hm_t = {}

        for i in range(len(t)):
            if t[i] in hm_t:
                hm_t[ t[i] ] += 1

            else:
                hm_t[  t[i] ] = 1
        
        hm_s = {}

        for i in hm_t:
            hm_s[ i ] = 0

        # sliding window
        l , r = 0 , 0
        res = s
        res_min = float("inf")

        for r in range( len(s) ):

            if s[r] in hm_s:
                hm_s[ s[r] ] += 1

            while all( hm_t[i] <= hm_s[i]  for i in hm_t ) :
                if r-l+1 < res_min:
                    res = s[l:r+1]
                    res_min = r-l+1

                if  s[l] in hm_s:
                    hm_s[ s[l] ] -= 1

                l += 1

        # 當 sliding window 沒有跑時
        if res_min == float("inf"):
            return ""

        return res