Interview question: three arrays and O(N*N)

Assume we have three arrays of length N which contain arbitrary numbers of type long. Then we are given a number M (of the same type) and our mission is to pick three numbers A, B and C one from each array (in other words A should be picked from first array, B from second one and C from third) so the sum A + B + C = M.

Question: could we pick all three numbers and end up with time complexity of O(N²)?

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Illustration:

Arrays are:

1) 6 5 8 3 9 2
2) 1 9 0 4 6 4
3) 7 8 1 5 4 3

And M we've been given is 19. Then our choice would be 8 from first, 4 from second and 7 from third.

This can be done in O(1) space and O(N²) time.

First lets solve a simpler problem:
Given two arrays A and B pick one element from each so that their sum is equal to given number K.

Sort both the arrays which takes O(NlogN).
Take pointers i and j so that i points to the start of the array A and j points to the end of B.
Find the sum A[i] + B[j] and compare it with K

• if A[i] + B[j] == K we have found the pair A[i] and B[j]
• if A[i] + B[j] < K, we need to increase the sum, so increment i.
• if A[i] + B[j] > K, we need to decrease the sum, so decrement j.

This process of finding the pair after sorting takes O(N).

Now lets take the original problem. We've got a third array now call it C.

So the algorithm now is :

foreach element x in C
  find a pair A[i], B[j] from A and B such that A[i] + B[j] = K - x
end for

The outer loop runs N times and for each run we do a O(N) operation making the entire algorithm O(N²).