Let’s say I have three arrays a, b, and c of equal length N. The elements of each of these arrays come from a totally ordered set, but are not sorted. I also have two index variables, i and j. For all i != j, I want to count the number of index pairs such that a[i] < a[j], b[i] > b[j] and c[i] < c[j]. Is there any way this can be done in less than O(N ^ 2) time complexity, for example by creative use of sorting algorithms?
Notes: The inspiration for this question is that, if you only have two arrays, a and b, you can find the number of index pairs such that a[i] < a[j] and b[i] > b[j] in O(N log N) with a merge sort. I’m basically looking for a generalization to three arrays.
For simplicity, you may assume that no two elements of any array are equal (no ties).
By sorting the array a and rearranging the arrays b and c at the same time, we can suppose that a[i] < a[j] <=> i < j. So we need to find the number of pairs (i,j) such that i < j, b[i] > b[j] and c[i] < c[j]. Let’s view (b[i], c[i]) as a point on a plane. We add the points one by one. Each time we add a point (b[j], c[j]), first we count the number of already added points (i < j) such that b[i] > b[j] and c[i] < c[j]. Then we add the point j and proceed to the next one. The sum of the numbers obtained at each step is our result.
Now it seems that this kind of queries can be fulfilled by two-dimensional segment tree: http://en.wikipedia.org/wiki/Segment_tree The cost of one iteration will be O(log^2 n), and the total complexity is O(n log^2 n).
(Note that I assume here that the elements of arrays are numbers. It’s OK, because using a sorting we can always replace the elements of an array with numbers from 1 to n so that the order was preserved.)
Edit: In fact, a simpler structure called Fenwick tree or binary indexed tree is sufficient. See this link: http://www.topcoder.com/tc?module=Static&d1=tutorials&d2=binaryIndexedTrees#2d