solve: minimumWindowSubstring

Author: GaggleHelena

minimum-window-substring/yolophg.py

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+# Time Complexity: O(N) - go through the string with two pointers, so it's basically O(N).
+# Space Complexity: O(1) - only storing character frequencies (max 52 keys for a-z & A-Z), so it's effectively constant space.
+
+class Solution:
+    def minWindow(self, s: str, t: str) -> str:
+        # store character counts for t
+        target_count = Counter(t)
+        # window character count
+        window_count = defaultdict(int)
+        
+        # start of the window
+        left = 0  
+        # min substring
+        min_substring = ""
+        # tracks how many characters match the required count
+        matched_chars = 0  
+        # unique characters needed
+        required_chars = len(target_count)  
+        
+        for right, char in enumerate(s):
+            # expand window by adding the rightmost character
+            if char in target_count:
+                window_count[char] += 1
+                if window_count[char] == target_count[char]:
+                    matched_chars += 1  
+
+            # try shrinking the window if all required characters are present
+            while matched_chars == required_chars:
+                # update min substring if this one is shorter
+                if min_substring == "" or (right - left + 1) < len(min_substring):
+                    min_substring = s[left:right + 1]
+
+                # remove leftmost character and move left pointer
+                if s[left] in window_count:
+                    window_count[s[left]] -= 1
+                    if window_count[s[left]] < target_count[s[left]]:
+                        matched_chars -= 1
+                left += 1  
+
+        return min_substring