Merge pull request #2405 from gyeo-ri/main

[gyeo-ri] WEEK 02 solutions

Author: SamTheKorean

longest-palindromic-substring/gyeo-ri.py

@@ -0,0 +1,70 @@
+"""
+[결과 요약]
+# 재시도횟수: 4회
+    1. 모든 문자열 조합을 계산하기(O(n^3))
+    2. 중심 확장(O(n^2)): 문자열을 순서대로 순회하며 중심을 기준으로 좌우 대칭 비교"""
+
+
+class Solution:
+    def longestPalindrome(self, s: str) -> str:
+        def _expand_around(left, right) -> tuple[int, int]:
+            while left >= 0 and right < length_s and s[left] == s[right]:
+                left -= 1
+                right += 1
+            return left + 1, right - 1
+
+        # 초깃값
+        length_s = len(s)
+        best_left, best_right = 0, 0
+
+        # 1. 중심을 한 칸씩 이동하면서 반복하기
+        for i in range(len(s)):
+            # 2. 짝수/홀수 각각에 대해 탐색하기
+            odd_left, odd_right = _expand_around(i, i)
+            even_left, even_right = _expand_around(i, i + 1)
+
+            best_length = best_right - best_left
+            odd_length = odd_right - odd_left
+            even_length = even_right - even_left
+
+            if odd_length > best_length:
+                best_left, best_right = odd_left, odd_right
+            if even_length > best_length:
+                best_left, best_right = even_left, even_right
+
+        return s[best_left : best_right + 1]
+
+
+"""
+#1. 브루트포스(O(n^3)): 순차 탐색하는 비효율적 알고리즘
+class Solution:
+    def longestPalindrome(self, s: str) -> str:
+        answer = ""
+
+        for i in range(len(s)):
+            for j in range(i, len(s)):
+                subset = s[i : j + 1]
+
+                if subset == subset[::-1]:
+                    if len(answer) < len(subset):
+                        answer = subset
+        return answer
+"""
+
+
+if __name__ == "__main__":
+    test_cases = [
+        ("babad", ["bab", "aba"]),
+        ("bb", ["bb"]),
+        ("aaabaaa", ["aaabaaa"]),
+        ("c", ["c"]),
+        ("123a24542a0321", ["a24542a"]),
+    ]
+
+    solution = Solution()
+    for idx, case_ in enumerate(test_cases):
+        s, answers = case_
+        result = solution.longestPalindrome(s)
+        assert (
+            result in answers
+        ), f"Test Case {idx} Failed: Expected one of {",".join(answers)}, Got {result}"