Minimum Window Substring

Author: anantkaushik

76-minimum-window-substring.py

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+"""
+Problem Link: https://leetcode.com/problems/minimum-window-substring/
+
+Given two strings s and t, return the minimum window in s which will contain all the characters in t. 
+If there is no such window in s that covers all characters in t, return the empty string "".
+Note that If there is such a window, it is guaranteed that there will always be only one unique minimum window in s.
+
+Example 1:
+Input: s = "ADOBECODEBANC", t = "ABC"
+Output: "BANC"
+
+Example 2:
+Input: s = "a", t = "a"
+Output: "a"
+ 
+Constraints:
+1 <= s.length, t.length <= 105
+s and t consist of English letters.
+
+Follow up: Could you find an algorithm that runs in O(n) time?
+"""
+class Solution:
+ def minWindow(self, s: str, t: str) -> str:
+ count = {}
+ for c in t:
+ count[c] = count.get(c, 0) + 1
+ 
+ # counter represents the number of chars of t to be found in s.
+ start, end, min_len, min_start, counter = 0, 0, float('inf'), 0, len(t)
+ 
+ while end < len(s):
+ 
+ # If char in s exists in t, decrease counter
+ if count.get(s[end], 0) > 0:
+ counter -= 1
+ 
+ # Decrease count[s[end]]. If char does not exist in t, count[s[end]] will be negative.
+ count[s[end]] = count.get(s[end], 0) - 1
+ 
+ end += 1
+ 
+ # When we found a valid window, move start to find smaller window.
+ while counter == 0:
+ 
+ if (end - start) < min_len:
+ min_len = end - start
+ min_start = start
+ 
+ count[s[start]] += 1
+ 
+ # When char exists in t, increase counter.
+ if count[s[start]] > 0:
+ counter += 1
+ 
+ start += 1
+ 
+ return s[min_start: min_start + min_len] if min_len != float('inf') else ""