Update minimum-window-substring.py

Python/minimum-window-substring.py

@@ -1,38 +1,44 @@
 # Time:  O(n)
 # Space: O(k), k is the number of different characters
-#
-# 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).
+
+# 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).
 # 
 # For example,
 # S = "ADOBECODEBANC"
 # T = "ABC"
 # Minimum window is "BANC".
 # 
 # Note:
-# If there is no such window in S that covers all characters in T, return the emtpy string "".
+# If there is no such window in S that covers all characters in T,
+# return the emtpy string "".
 # 
-# If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.
-#
+# If there are multiple such windows, you are guaranteed that
+# there will always be only one unique minimum window in S.
 
-class Solution:
-    # @return a string
-    def minWindow(self, S, T):
+class Solution(object):
+    def minWindow(self, s, t):
+        """
+        :type s: str
+        :type t: str
+        :rtype: str
+        """
         current_count = [0 for i in xrange(52)]
         expected_count = [0 for i in xrange(52)]
 
-        for char in T:
+        for char in t:
             expected_count[ord(char) - ord('a')] += 1
 
         i, count, start, min_width, min_start = 0, 0, 0, float("inf"), 0
-        while i < len(S):
-            current_count[ord(S[i]) - ord('a')] += 1
-            if current_count[ord(S[i]) - ord('a')] <= expected_count[ord(S[i]) - ord('a')]:
+        while i < len(s):
+            current_count[ord(s[i]) - ord('a')] += 1
+            if current_count[ord(s[i]) - ord('a')] <= expected_count[ord(s[i]) - ord('a')]:
                 count += 1
 
-            if count == len(T):
-                while expected_count[ord(S[start]) - ord('a')] == 0 or\
-                      current_count[ord(S[start]) - ord('a')] > expected_count[ord(S[start]) - ord('a')]:
-                    current_count[ord(S[start]) - ord('a')] -= 1
+            if count == len(t):
+                while expected_count[ord(s[start]) - ord('a')] == 0 or\
+                      current_count[ord(s[start]) - ord('a')] > expected_count[ord(s[start]) - ord('a')]:
+                    current_count[ord(s[start]) - ord('a')] -= 1
                     start += 1
 
                 if min_width > i - start + 1:
@@ -43,7 +49,8 @@ def minWindow(self, S, T):
         if min_width == float("inf"):
             return ""
 
-        return S[min_start:min_start + min_width]
+        return s[min_start:min_start + min_width]
+
 
 if __name__ == "__main__":
     print Solution().minWindow("ADOBECODEBANC", "ABC")