Today I wanted to solve this problem: Heavy Intervals But I got TLE3 code: 260013398. that's why I changed the code a little bit and my code(260014069) was correct. But I don't understand how? Can you explain?
# | User | Rating |
---|---|---|
1 | jiangly | 4039 |
2 | tourist | 3841 |
3 | jqdai0815 | 3682 |
4 | ksun48 | 3590 |
5 | ecnerwala | 3542 |
6 | Benq | 3535 |
7 | orzdevinwang | 3526 |
8 | gamegame | 3477 |
9 | heuristica | 3357 |
10 | Radewoosh | 3355 |
# | User | Contrib. |
---|---|---|
1 | cry | 169 |
2 | -is-this-fft- | 165 |
3 | atcoder_official | 160 |
3 | Um_nik | 160 |
5 | djm03178 | 158 |
6 | Dominater069 | 156 |
7 | adamant | 153 |
8 | luogu_official | 152 |
9 | awoo | 151 |
10 | TheScrasse | 147 |
Today I wanted to solve this problem: Heavy Intervals But I got TLE3 code: 260013398. that's why I changed the code a little bit and my code(260014069) was correct. But I don't understand how? Can you explain?
Name |
---|
b.upperbound takes $$$O(logn)$$$ while upperbound(all(b), a[i]) takes $$$O(n)$$$ time.
for sets , b.upper_bound is O(log(n)) but upper_bound(all(b)) is O(n)
Using set.upper_bound() is O(log n), while std::upper_bound(set.begin(), set.end(), value) adds O(n) complexity due to linear search, potentially causing time limit.