Thanks for the explanation.

Thanks.

Problem C: I did the same n! — 2^(n-1) but got wrong answer on pretest 6. Can anyone tell me what's wrong? Here is the link to my solution

Nope. AFAIK you cannot create a hash sum array for this question. You have to create a freq array of numbers and then run a loop from (2 to 2*n) and then search on the freq array for the sum, as given in the editorial.

Yes, you have to select the weight with the maximum frequency. But the way you are creating the sum array is wrong because that loop is running n^2 times and creating almost n^2 pairs. The problem is that when you hash them to the sum variable, you are using one person more than once (in different pars), which leads to a wrong sum array.

In the first for loop you are doing sum[a[i]+a[j]]++ which is wrong because lets say there are 3 people with weights {1,3,3} In the first iteration , you are using the zeroeth person thrice (when (i,j) is (0,0) (0,1) (0,2)). In the first iteration itself sum[4] will be 2 which is not possible since if the zeroeth person made a team with the first person for sum=4, he cannot make another team with the second person. As a person cannot be in 2 teams, this approach fails.

Ohhhhh... Got it now Thanks a lot.

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