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Add detailed explanation and implementation for the Top K Frequent Elements problem, including various approaches such as using collections.Counter, heapq, and bucket sort.
mamo3gr
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Feb 2, 2026
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| frequency = {} | ||
| for number in nums: | ||
| frequency[number] = frequency.get(number, 0) + 1 |
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[fyi]
defaultdict を使うとさらにシンプルに書けます。
Suggested change
| frequency = {} | |
| for number in nums: | |
| frequency[number] = frequency.get(number, 0) + 1 | |
| frequency = defaultdict(int) | |
| for number in nums: | |
| frequency[number] += 1 |
mamo3gr
reviewed
Feb 2, 2026
Comment on lines
+60
to
+62
| top_k_items = sorted(frequency.items(), key=itemgetter(1), reverse=True)[:k] | ||
| top_k_numbers = list(dict(top_k_items)) | ||
| return top_k_numbers |
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こちらも、もうちょっとシンプルに書けます。
Suggested change
| top_k_items = sorted(frequency.items(), key=itemgetter(1), reverse=True)[:k] | |
| top_k_numbers = list(dict(top_k_items)) | |
| return top_k_numbers | |
| top_k_numbers = sorted(frequency, key=frequency.get, reverse=True)[:k] | |
| return top_k_numbers |
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今回の問題: 347. Top K Frequent Elements
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