Question : RRFRNDS I used brute force to solve this and was expecting a TLE but got a WA instead Somebody provide me some hints to solve this
| # | User | Rating |
|---|---|---|
| 1 | Benq | 3857 |
| 2 | jiangly | 3810 |
| 3 | maroonrk | 3534 |
| 4 | tourist | 3528 |
| 5 | Kevin114514 | 3510 |
| 6 | turmax | 3411 |
| 7 | Um_nik | 3387 |
| 8 | Radewoosh | 3367 |
| 9 | heuristica | 3322 |
| 10 | strapple | 3317 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 157 |
| 2 | maspy | 150 |
| 3 | Um_nik | 145 |
| 4 | Errichto | 139 |
| 5 | adamant | 136 |
| 6 | maroonrk | 134 |
| 7 | nik_exists | 133 |
| 7 | DNR | 133 |
| 9 | AmShZ | 130 |
| 10 | Dominater069 | 129 |
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You can find editorial at codechef. Editorial link has to be under the statement as i remember.
Let's call list[i] is all friend of user[i], after that, we check all pair (i, j), if currently i and j is not friend and exists an user in list[i] is also friend of j, then pair (i, j) is valid, we increase the answer to 1.
That's an O(N^3) implementation and will time out