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t9a-dev
reviewed
Dec 5, 2025
| return top_k_frequenct_elements | ||
| ``` | ||
| https://github.com/potrue/leetcode/pull/9#discussion_r2083523591 | ||
| 以上を参考に, bucketの初期化部分を以下のように変更したところ3ms程度まで短縮できた. |
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今回のケースには当てはまらないかも知れませんが、可読性を犠牲に速度を求めるのかという視点もあります。
t0hsumi/leetcode#1 (comment)
olsen-blue/Arai60#34 (comment)
nodchip
reviewed
Dec 5, 2025
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| sorted_number_to_count = dict(sorted(number_to_count.items(), reverse=True, key=lambda item: item[1])) | ||
| numbers_sorted_by_count = list(sorted_number_to_count.keys()) | ||
| return [numbers_sorted_by_count[i] for i in range(k)] |
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return numbers_sorted_by_count[:k]のほうがシンプルだと思いました。
Owner
Author
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ありがとうございます
何故か複雑な書き方をしてました…
| ```py | ||
| class Solution: | ||
| def topKFrequent(self, nums: List[int], k: int) -> List[int]: | ||
| number_to_count = {} |
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defaultdict を用いることで、シンプルにすることができると思います。
number_to_count = defaultdict(int)
for num in nums:
number_to_count[num] += 1
TakayaShirai
reviewed
Dec 8, 2025
Comment on lines
+51
to
+52
| for number, frequency in number_to_frequency.items(): | ||
| heapq.heappush(heap, (-frequency, number)) |
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状況にもよると思いますが、nums の要素数に比べて k がかなり小さい場合などは、以下のように k 個の要素のみをヒープに残す方が、 push や pop の操作で内部でヒープを更新するのに必要な処理を減らせて良さそうです。
for number, frequency in number_to_frequency.items():
if len(heap) < k:
heapq.heappush(heap, (frequency, number))
continue
if frequency > heap[0][0]:
heapq.heappushpop(heap, (frequency, number))
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解く問題
Top K Frequent Elements
次に解く問題
Find K Pairs With Smallest Sums