Longest Palindrome

0409.Longest-Palindrome/memo.md

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+# 409. Longest Palindrome
+
+## step1
+解くだけなら 5分ぐらい
+
+## step2
+
+Counterで書き直す、変数名の改善
+
+> 正攻法でないものとして、Manacher's algorithm というのがあります。
+
+別の問題の解法
+
+https://leetcode.com/problems/longest-palindromic-substring/description/?envType=problem-list-v2&envId=rab78cw1
+
+(忘れていたので書き直していても良いかもしれない)
+
+https://github.com/Kitaken0107/GrindEasy/pull/27#discussion_r1656471861
+
+https://github.com/huyfififi/coding-challenges/pull/17
+
+https://github.com/NobukiFukui/Grind75-ProgrammingTraining/pull/31
+
+https://github.com/ryosuketc/leetcode_grind75/pull/17
+
+おおむね同じ解法だが 1 pathの書き方を真似しておく。 最後は `len(s) - odd_count + 1`としている。

0409.Longest-Palindrome/step1.py

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+class Solution:
+    def longestPalindrome(self, s: str) -> int:
+        char_to_count = {}
+        for c in s:
+            char_to_count[c] = char_to_count.setdefault(c, 0) + 1
+
+        exist_odd_count = False
+        half_length = 0
+        for c, count in char_to_count.items():
+            half_length += count // 2
+            if count % 2 == 1:
+                exist_odd_count = True
+
+        return half_length * 2 if not exist_odd_count else half_length * 2 + 1

0409.Longest-Palindrome/step2.py

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+from collections import Counter
+
+
+class Solution:
+    def longestPalindrome(self, s: str) -> int:
+        counts = Counter(s)
+        length = 0
+        has_odd_count = False
+
+        for count in counts.values():
+            length += (count // 2) * 2
+            if count % 2 == 1:
+                has_odd_count = True
+
+        return length + 1 if has_odd_count else length
+
+
+from collections import defaultdict
+
+
+class Solution:
+    def longestPalindrome(self, s: str) -> int:
+        char_to_count = defaultdict(int)
+        odd_count = 0
+
+        for c in s:
+            char_to_count[c] += 1
+            if char_to_count[c] % 2 == 1:
+                odd_count += 1
+            else:
+                odd_count -= 1
+
+        if odd_count <= 1:
+            return len(s)
+        return len(s) - odd_count + 1