(leetcode)Solution for minimum window substring

Author: pratikj697

Leetcode - Top Interview Questions/minimumwindowsubstring.py

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+def minWindow(self, s, t):
+    """
+    :type s: str
+    :type t: str
+    :rtype: str
+    """
+
+    if not t or not s:
+        return ""
+
+    # Dictionary which keeps a count of all the unique characters in t.
+    dict_t = Counter(t)
+
+    # Number of unique characters in t, which need to be present in the desired window.
+    required = len(dict_t)
+
+    # left and right pointer
+    l, r = 0, 0
+
+    # formed is used to keep track of how many unique characters in t are present in the current window in its desired frequency.
+    # e.g. if t is "AABC" then the window must have two A's, one B and one C. Thus formed would be = 3 when all these conditions are met.
+    formed = 0
+
+    # Dictionary which keeps a count of all the unique characters in the current window.
+    window_counts = {}
+
+    # ans tuple of the form (window length, left, right)
+    ans = float("inf"), None, None
+
+    while r < len(s):
+
+        # Add one character from the right to the window
+        character = s[r]
+        window_counts[character] = window_counts.get(character, 0) + 1
+
+        # If the frequency of the current character added equals to the desired count in t then increment the formed count by 1.
+        if character in dict_t and window_counts[character] == dict_t[character]:
+            formed += 1
+
+        # Try and contract the window till the point where it ceases to be 'desirable'.
+        while l <= r and formed == required:
+            character = s[l]
+
+            # Save the smallest window until now.
+            if r - l + 1 < ans[0]:
+                ans = (r - l + 1, l, r)
+
+            # The character at the position pointed by the `left` pointer is no longer a part of the window.
+            window_counts[character] -= 1
+            if character in dict_t and window_counts[character] < dict_t[character]:
+                formed -= 1