feat: solve DaleStudy#285 with python

minimum-window-substring/EGON.py

@@ -0,0 +1,67 @@
+from collections import Counter
+from typing import List
+from unittest import TestCase, main
+
+
+class Solution:
+    def minWindow(self, s: str, t: str) -> str:
+        return self.solve_two_pointer(s, t)
+
+    """
+    Runtime: 129 ms (Beats 50.44%)
+    Time Complexity: O(S)
+        - 문자열 s를 enumerate로 순회하는데 O(S)
+        - 순회 후 left를 갱신하는 while문에서 left가 0부터 n까지 단조증가하므로 총 조회는 O(S)
+        > O(S) + O(S) ~= O(S)
+
+    Memory: 17.32 MB (Beats 32.52%)
+    Space Complexity: O(S)
+        - counter 변수의 초기 크기는 O(T)
+            - 반복문을 조회하며 counter 갱신, 최악의 경우 s의 모든 문자가 다르고 s == t인 경우 이므로 O(S), upper bound
+        > O(S)
+    """
+    def solve_two_pointer(self, s: str, t: str) -> str:
+        counter = Counter(t)
+        missing = len(t)
+        left = start = end = 0
+        for right, char in enumerate(s, start=1):
+            missing -= counter[char] > 0
+            counter[char] -= 1
+
+            if missing == 0:
+                while left < right and counter[s[left]] < 0:
+                    counter[s[left]] += 1
+                    left += 1
+
+                if not end or right - left <= end - start:
+                    start, end = left, right
+
+                counter[s[left]] += 1
+                missing += 1
+                left += 1
+
+        return s[start:end]
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        s = "ADOBECODEBANC"
+        t = "ABC"
+        output = "BANC"
+        self.assertEqual(Solution.minWindow(Solution(), s, t), output)
+
+    def test_2(self):
+        s = "a"
+        t = "a"
+        output = "a"
+        self.assertEqual(Solution.minWindow(Solution(), s, t), output)
+
+    def test_3(self):
+        s = "a"
+        t = "aa"
+        output = ""
+        self.assertEqual(Solution.minWindow(Solution(), s, t), output)
+
+
+if __name__ == '__main__':
+    main()