3sum在n ^ 2 * log（n）时间中有4个数字的变体？

Question

I am working on a project for an algorithms course I'm in and I'm completely at an impass. The assignment is to find all sets of 4 numbers in an array for which i+j=k+l in O(n^2*log(n)) time.

I know this is similar to the 3sum problem, where you have to find all the sets in an array for which i+j+k=0. We have discussed a solution to this problem in our lecture that solves it in O(n^2*log(n)) time by iterating through all unique pairs of 2 (n^2 time) and using a binary search on a sorted array to find a value that satisfies the problem (log(n) time).

However, I don't see how the problem with 4 numbers could be solved in this time. I think it's a safe guess that the log(n) in the complexity comes from a binary search, which i'll use for the last one. However, that would mean I have to iterate over all possible combinations of 3 in n^2 time, which I just don't think is possible.

I'm not looking for someone to do my work for me, but maybe if someone knows how this can be solved, they could give me a gentle tap in the right direction. Thanks.

Edit: it might also be helpful to note that I am required to solve this using sorting. Once I do that, I have to write another implementation that makes it faster using hashing, but I think I'll be able to do that just fine on my own. I just need to figure out how I can solve the problem with sorting first.

Answer 1

Keep a sorted list of sums and pairs. For instance, given the list

[1, 2, 4, 8, 16, 32, 9, 82]

we will want to identify 1+9 = 2+8

Identify the largest two numbers in the list, O**(N)**. In this case, they're (82, 32), a sum of 114. Allocate an array pair_sum of 114 locations; set all locations to null pointers.

Now iterate through the list for your i, j pairs. For each pair, insert the two numbers as a tuple-value at index i+j. When you get a collision at some index, you're done: you found a second pair-sum.

I'll outline this in some not-quite-pseudo code; you can translate to your favorite implementation language.

bank = [1, 2, 4, 8, 16, 32, 9, 82] size = length(bank)

// Find the two largest numbers and allocate the array ...
pair_sum = array[114], initialize to all nulls

for lo in 0:size-1
    for hi in lo+1:size
        sum = bank[lo] + bank[hi]
        if pair_sum[sum]
            // There is already a pair with that sum
            print pair_sum[sum], "and", bank[lo], bank[hi]
        else
            // Record new sum pair
            pair_sum[sum] = tuple(bank[lo], bank[hi])

This is O(N^2) , with bounded space dependent on the array values.

If you aren't allowed to use the sum as an index for practical reasons, I think you can adapt this to a binary search and insertion, giving you the log(n) component.