242. Valid Anagram #65
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eito2002
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Nov 27, 2025
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| class Solution: | ||
| def isAnagram(self, s: str, t: str) -> bool: | ||
| if len(s) != len(t): | ||
| return False | ||
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| s_char_to_count = {} | ||
| for ch in s: | ||
| if ch not in s_char_to_count: | ||
| s_char_to_count[ch] = 0 | ||
| s_char_to_count[ch] += 1 | ||
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| remaining_count = s_char_to_count | ||
| for ch in t: | ||
| if remaining_count.get(ch, 0) == 0: | ||
| return False | ||
| remaining_count[ch] -= 1 | ||
| return True |
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remaining_countとs_char_to_countは同じオブジェクトを指しているので、remaining_countは無くても良いかなと思いました
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ありがとうございます。
s_char_to_countがtのカウントにより減少していくのは不自然かなと思いつつ、わざわざ複製するほどでもないので同じオブジェクト参照のエイリアス変数を定義しましたが、確かに中途半端でかえって分かりにくい気がします。
おっしゃるとおりremaining_countなしの方が良いですね。
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242. Valid Anagram
Next: 235. Lowest Common Ancestor of a Binary Search Tree