| # | User | Rating |
|---|---|---|
| 1 | Benq | 3676 |
| 2 | Kevin114514 | 3659 |
| 3 | jiangly | 3640 |
| 4 | turmax | 3450 |
| 5 | tourist | 3408 |
| 6 | strapple | 3360 |
| 7 | Um_nik | 3354 |
| 8 | squareOf105 | 3348 |
| 9 | maroonrk | 3312 |
| 10 | idontwannadothis…e | 3191 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 154 |
| 2 | nik_exists | 151 |
| 3 | maspy | 149 |
| 4 | AmShZ | 146 |
| 5 | Um_nik | 143 |
| 6 | Errichto | 139 |
| 7 | BledDest | 136 |
| 8 | adamant | 135 |
| 9 | maroonrk | 133 |
| 10 | cry | 130 |
|
0
That is my AC Code using deque (python): Code |
|
0
I think problem B can be solved by Brute-Force and Deque data structure. Because the bottom card removal operation can cost O(n)(remove arr[0] in array ) and repeating it many times like that can make the complexity O(n^2) and it will be TLE. We can solve this problem by Deque which makes the bottom card removal only have complexity O(n) and helps the complexity of the code from O(n^2) to O(n) This is just a suggestion on the solution of problem B improved from Brute-Force, please do not downvote me |
| Name |
|---|


