703. Kth Largest Element in a Stream #9
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sakzk
reviewed
Jun 13, 2024
| - 最小ヒープを使う | ||
| - ヒープのサイズをkに保つ | ||
| - 初期化: O(n log k), 各クエリ: O(log k) | ||
| - fenwick tree上の二分探索(もしくはsegment tree上の二分探索) |
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(コードへのコメントではないですが。) fenwick tree 知らなかったので勉強になりました！
TORUS0818
reviewed
Jun 13, 2024
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| for num in nums: | ||
| if len(self.num_min_heap) < k: | ||
| heapq.heappush(self.num_min_heap, num) | ||
| else: | ||
| heapq.heappushpop(self.num_min_heap, num) |
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こちら、addを呼ぶといいと思います。
for num in nums:
self.add(num)
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気付きませんでした。シンプルになりますね。
ありがとうございます。
oda
reviewed
Jun 13, 2024
| - kが各クエリで変わるときは、この方法になると思う(今回はk固定なのでスキップ) | ||
| - 初期化: O(n log n), 各クエリ: O(log n) | ||
| - 平衡二分探索木 | ||
| - たぶん最終手段だと思う |
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Python で使いたかったら、標準でないライブラリーがあります。C++ では標準で平衡木がありますね。
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こちら調べてみました。
- C++標準のmultisetを使うと、前から辿る必要があるので、kth elementをO(k)で取得可能
- g++拡張の__gnu_pbdsにあるtreeを使うと、各ノードにsub treeのサイズを持たせられるオプションがあり、kth elemntをO(log n)で取得可能(LeetCodeでも使えました)
- PythonでもLeetCodeで使えるものを探してみたところ、sortedcontainerのSortedListを使うとO(log n)でkth elementを取得できた
- こちらは内部実装を確認したところおそらく平衡木ベースではなさそう
- https://grantjenks.com/docs/sortedcontainers/sortedlist.html
実装を追加してます。-> f382024
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問題へのリンク
https://leetcode.com/problems/kth-largest-element-in-a-stream/description/
次に解く問題
347. Top K Frequent Elements
README.mdへ頭の中の言語化と記録をしています。