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
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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