set<int>s;
s.erase(s.begin());
I wanted to know whether removing the minimum element in set is a O(logN) or O(1) operation ?
I am quite confused about this situation.
Thanks in advance..
→ Обратите внимание
Соревнование идет
2023 Post World Finals Online ICPC Challenge powered by Huawei
10 дней
Зарегистрироваться »
2023 Post World Finals Online ICPC Challenge powered by Huawei
10 дней
Зарегистрироваться »
→ Трансляции
→ Лидеры (рейтинг)
| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | tourist | 3690 |
| 2 | jiangly | 3647 |
| 3 | Benq | 3581 |
| 4 | orzdevinwang | 3570 |
| 5 | Geothermal | 3569 |
| 5 | cnnfls_csy | 3569 |
| 7 | Radewoosh | 3509 |
| 8 | ecnerwala | 3486 |
| 9 | jqdai0815 | 3474 |
| 10 | gyh20 | 3447 |
| Страны | Города | Организации | Всё → |
→ Лидеры (вклад)
| № | Пользователь | Вклад |
|---|---|---|
| 1 | maomao90 | 174 |
| 2 | awoo | 164 |
| 3 | adamant | 162 |
| 4 | TheScrasse | 160 |
| 5 | nor | 158 |
| 6 | maroonrk | 156 |
| 7 | -is-this-fft- | 152 |
| 8 | orz | 146 |
| 9 | pajenegod | 145 |
| 9 | SecondThread | 145 |
→ Найти пользователя
→ Прямой эфир
Codeforces (c) Copyright 2010-2024 Михаил Мирзаянов
Соревнования по программированию 2.0
Время на сервере: 10.05.2024 14:23:49 (i2).
Десктопная версия, переключиться на мобильную.
При поддержке
It takes amortized constant time. See here set::begin and here set::erase.
Thanks buddy
It says just "constant" (doesn't say about amortisation), but I doubt that's true.
"For the first version (erase(position)), amortized constant." (where position is an iterator) from http://www.cplusplus.com/reference/set/set/erase/
ah, sorry, I just followed Jakube's links, but both of them apparently go to std::set::begin
Yes, I noticed that too :)
Sorry, I fixed the link.