Adding Efficient Solution for Problem - 76 - Minimum Window Substring

Author: Vanditg

Minimum Window Substring/EfficientSolution.py

@@ -0,0 +1,33 @@
+##==================================
+## Leetcode
+## Student: Vandit Jyotindra Gajjar
+## Year: 2020
+## Problem: 76
+## Problem Name: Minimum Window Substring      
+##===================================
+#
+#Given a string S and a string T, 
+#find the minimum window in S which will contain all the characters in T in complexity O(n).
+#
+#Example:
+#
+#Input: S = "ADOBECODEBANC", T = "ABC"
+#Output: "BANC"
+import collections 
+class Solution:
+    def minWindow(self, s, t):
+        tmp = collections.Counter(t)	#Initilaize tmp dictionary > key = letter and value = count
+        left, right, i, j, minStr = 0, 0, 0, 0, len(t)	#Initilaize left, right, i, j, and minStr
+        while right < len(s):	#Loop while right is less than the length of string 
+            if tmp[s[right]] > 0:	#Condition-check: If string's right value is greater than zero
+                minStr -= 1	#Update minStr by decreasing 1
+            tmp[s[right]] -= 1	#Update dictionary value for particular key by decreasing 1
+            right += 1	#Update right by increasing 1 
+            while minStr == 0:	#Loop while minStr is zero
+                if j == 0 or right - left < j - i:	#Condition-check: If j is zero or right - left value is less than j - i value
+                    i, j = left, right	#Update i, j by left, right
+                tmp[s[left]] += 1	#Update dictionary value for particular key by increasing 1
+                if tmp[s[left]] > 0:	#If dictionary value for particular key is greater than 0
+                    minStr += 1	#Update minStr by increasing 1
+                left += 1	#Update left by increasing 1
+        return s[i:j]	#Finally return s from i to j index