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 | tourist | 4009 |
2 | jiangly | 3773 |
3 | Radewoosh | 3646 |
4 | ecnerwala | 3624 |
5 | jqdai0815 | 3620 |
5 | Benq | 3620 |
7 | orzdevinwang | 3612 |
8 | Geothermal | 3569 |
8 | cnnfls_csy | 3569 |
10 | Um_nik | 3396 |
# | User | Contrib. |
---|---|---|
1 | Um_nik | 163 |
2 | cry | 161 |
3 | maomao90 | 160 |
4 | -is-this-fft- | 159 |
5 | awoo | 158 |
6 | atcoder_official | 157 |
7 | adamant | 155 |
7 | nor | 155 |
9 | maroonrk | 152 |
10 | Dominater069 | 148 |
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.