Three nearby numbers in three arrays

Question

Given three sorted floating-point arrays a[], b[], and c[], design a linearithmic algorithm to find three integers i,

Answer 1

Let us look at some potential ordering of the elements:

a[i] < b[j] < c[k]

Then we can see the following claim holds:

Target = |a[i] - b[j]| + |b[j] - c[k]| + |c[k] - a[i]|
       = b[j] - a[i] + c[k] - b[j] + c[k] - a[i]
       = 2 * (c[k] - a[i])

So, for any possible ordering, this is the minimization of the difference between two elements in two different arrays. So, simply minimize for every possible combination (a and b, b and c, c and a) as shown in the question you gave a reference to (can be done in linearithmic time for each pair).

Once you found the minimization for a pair, finding the matching element from the third array should be quite easy - simply go over that array and check each element.

Answer 2

The following algorithm is almost like merging three sorted arrays into one sorted one.

Keep one pointer for each array (i,j,k for A, B and C respectively). Initialize them to 0.

Then compute the difference between A[i], B[j], C[k] and update the lowest value achieved till now if necessary.

Increment the index in the array for which

array[index] =  min(A[i], B[j] and C[k]) 

if it has not reached the end.

That is:

If ( A[i] is the least ), then increment i.
else If ( B[j] is the least ), then increment j.
else increment k.

Keep doing the above till any one index runs past the end or you find a situation where all three A[i], B[j] and C[k] are equal.

EDIT:
If there are two duplicate candidates (say A[i] == B[j]), then increment both i and j. See for yourself why.

Also, if A[i+1] == A[i], then simply increment i again.
End Edit:

The above algorithm has O(N) time complexity.

Proof of correctness:
As shown by other answer, the difference depends only on two extremes of A[i], B[j], C[k].

So if A[i] < B[j] < C[k], then difference = 2*(C[k] - A[i]). Hence if we increment either j or k, then the difference can only increase. Hence we increment i.