i hv tried this question with naive ....
but i think it can be done with the help of dp...
any help??
→ Pay attention
Before contest
CodeTON Round 9 (Div. 1 + Div. 2, Rated, Prizes!)
24:07:28
Register now »
CodeTON Round 9 (Div. 1 + Div. 2, Rated, Prizes!)
24:07:28
Register now »
*has extra registration
→ Top rated
| # | User | Rating |
|---|---|---|
| 1 | tourist | 4009 |
| 2 | jiangly | 3823 |
| 3 | Benq | 3738 |
| 4 | Radewoosh | 3633 |
| 5 | jqdai0815 | 3620 |
| 6 | orzdevinwang | 3529 |
| 7 | ecnerwala | 3446 |
| 8 | Um_nik | 3396 |
| 9 | ksun48 | 3390 |
| 10 | gamegame | 3386 |
→ Top contributors
| # | User | Contrib. |
|---|---|---|
| 1 | cry | 167 |
| 2 | Um_nik | 163 |
| 3 | maomao90 | 162 |
| 3 | atcoder_official | 162 |
| 5 | adamant | 159 |
| 6 | -is-this-fft- | 158 |
| 7 | awoo | 157 |
| 8 | TheScrasse | 154 |
| 9 | Dominater069 | 153 |
| 9 | nor | 153 |
→ Find user
→ Recent actions
Codeforces (c) Copyright 2010-2024 Mike Mirzayanov
The only programming contests Web 2.0 platform
Server time: Nov/22/2024 17:27:32 (l1).
Desktop version, switch to mobile version.
Supported by
Yes, it can be done in O(N) with prefix/suffix sums. Let gi be the amount of green squares in the prefix [1, i] and ri be the amount of red squares in the suffix [i, N]. Then the answer is the minimum gi + ri + 1.