Question

Is there a more efficient way to achieve this:
Given an array A of size n and two positive integers a and b, find the sum floor(abs(A[i]-A[j])*a/b) taken over all pairs (i, j) where 0 <= i < j < n.

int A[n];
int a, b; // assigned some positive integer values
...
int total = 0;
for (int i = 0; i < n; i++) {
    for (int j = i+1; j < n; j++) {
        total += abs(A[i]-A[j])*a/b; // want integer division here
    }
}

To optimize this a little bit, I sorted the array (O(nlogn)) and then didn’t use an abs function. Also, I cached the value a[i] before the inner for loop, so I could just read stuff from A sequentially. I was considering precomputing a/b and storing that in a float, but the extra casting just makes it slower (especially since I want to take the floor of the result).

I couldn’t come up with a solution that was better than O(n^2).

Answer 1

Yes, there is a more efficient algorithm. It can be done it in O(n*log n). I don’t expect there to be an asymptotically faster way, but I’m far from any idea of a proof.

Algorithm

First sort the array in O(n*log n) time.

Now, let us look at the terms

floor((A[j]-A[i])*a/b) = floor ((A[j]*a - A[i]*a)/b)

for 0 <= i < j < n. For each 0 <= k < n, write A[k]*a = q[k]*b + r[k] with 0 <= r[k] < b.

For A[k] >= 0, we have q[k] = (A[k]*a)/b and r[k] = (A[k]*a)%b with integer division, for A[k] < 0, we have q[k] = (A[k]*a)/b - 1 and r[k] = b + (A[k]*a)%b unless b divides A[k]*a, in which case we have q[k] = (A[k]*a)/b and r[k] = 0.

Now we rewrite the terms:

floor((A[j]*a - A[i]*a)/b) = floor(q[j] - q[i] + (r[j] - r[i])/b)
                           = q[j] - q[i] + floor((r[j] - r[i])/b)

Each q[k] appears k times with positive sign (for i = 0, 1, .. , k-1) and n-1-k times with negative sign (for j = k+1, k+2, ..., n-1), so its total contribution to the sum is

(k - (n-1-k))*q[k] = (2*k+1-n)*q[k]

The remainders still have to be accounted for. Now, since 0 <= r[k] < b, we have

-b < r[j] - r[i] < b

and floor((r[j]-r[i])/b) is 0 when r[j] >= r[i] and -1 when r[j] < r[i]. So

                            n-1
 ∑ floor((A[j]-A[i])*a/b) =  ∑ (2*k+1-n)*q[k] - inversions(r)
i<j                         k=0

where an inversion is a pair (i,j) of indices with 0 <= i < j < n and r[j] < r[i].

Calculating the q[k] and r[k] and summing the (2*k+1-n)*q[k] is done in O(n) time.

It remains to efficiently count the inversions of the r[k] array.

For each index 0 <= k < n, let c(k) be the number of i < k such that r[k] < r[i], i.e. the number of inversions in which k appears as the larger index.

Then obviously the number of inversions is ∑ c(k).

On the other hand, c(k) is the number of elements that are moved behind r[k] in a stable sort (stability is important here).

Counting these moves, and hence the inversions of an array is easy to do while merge-sorting it.

Thus the inversions can be counted in O(n*log n) too, giving an overall complexity of O(n*log n).

Code

A sample implementation with a simple unscientific benchmark (but the difference between the naive quadratic algorithm and the above is so large that an unscientific benchmark is conclusive enough).

#include <stdlib.h>
#include <stdio.h>
#include <time.h>

long long mergesort(int *arr, unsigned elems);
long long merge(int *arr, unsigned elems, int *scratch);
long long nosort(int *arr, unsigned elems, long long a, long long b);
long long withsort(int *arr, unsigned elems, long long a, long long b);

int main(int argc, char *argv[]) {
    unsigned count = (argc > 1) ? strtoul(argv[1],NULL,0) : 1000;
    srand(time(NULL)+count);
    long long a, b;
    b = 1000 + 9000.0*rand()/(RAND_MAX+1.0);
    a = b/3 + (b-b/3)*1.0*rand()/(RAND_MAX + 1.0);
    int *arr1, *arr2;
    arr1 = malloc(count*sizeof *arr1);
    arr2 = malloc(count*sizeof *arr2);
    if (!arr1 || !arr2) {
        fprintf(stderr,"Allocation failed\n");
        exit(EXIT_FAILURE);
    }
    unsigned i;
    for(i = 0; i < count; ++i) {
        arr1[i] = 20000.0*rand()/(RAND_MAX + 1.0) - 2000;
    }
    for(i = 0; i < count; ++i) {
        arr2[i] = arr1[i];
    }
    long long res1, res2;
    double start = clock();
    res1 = nosort(arr1,count,a,b);
    double stop = clock();
    printf("Naive:   %lld in %.3fs\n",res1,(stop-start)/CLOCKS_PER_SEC);
    start = clock();
    res2 = withsort(arr2,count,a,b);
    stop = clock();
    printf("Sorting: %lld in %.3fs\n",res2,(stop-start)/CLOCKS_PER_SEC);
    return EXIT_SUCCESS;
}

long long nosort(int *arr, unsigned elems, long long a, long long b) {
    long long total = 0;
    unsigned i, j;
    long long m;
    for(i = 0; i < elems-1; ++i) {
        m = arr[i];
        for(j = i+1; j < elems; ++j) {
            long long d = (arr[j] < m) ? (m-arr[j]) : (arr[j]-m);
            total += (d*a)/b;
        }
    }
    return total;
}

long long withsort(int *arr, unsigned elems, long long a, long long b) {
    long long total = 0;
    unsigned i;
    mergesort(arr,elems);
    for(i = 0; i < elems; ++i) {
        long long q, r;
        q = (arr[i]*a)/b;
        r = (arr[i]*a)%b;
        if (r < 0) {
            r += b;
            q -= 1;
        }
        total += (2*i+1LL-elems)*q;
        arr[i] = (int)r;
    }
    total -= mergesort(arr,elems);
    return total;
}

long long mergesort(int *arr, unsigned elems) {
    if (elems < 2) return 0;
    int *scratch = malloc((elems + 1)/2*sizeof *scratch);
    if (!scratch) {
        fprintf(stderr,"Alloc failure\n");
        exit(EXIT_FAILURE);
    }
    return merge(arr, elems, scratch);
}

long long merge(int *arr, unsigned elems, int *scratch) {
    if (elems < 2) return 0;
    unsigned left = (elems + 1)/2, right = elems-left, i, j, k;
    long long inversions = 0;
    inversions += merge(arr, left, scratch);
    inversions += merge(arr+left,right,scratch);
    if (arr[left] < arr[left-1]) {
        for(i = 0; i < left; ++i) {
            scratch[i] = arr[i];
        }
        i = 0; j = 0; k = 0;
        int *lptr = scratch, *rptr = arr+left;
        while(i < left && j < right) {
            if (rptr[j] < lptr[i]) {
                arr[k++] = rptr[j++];
                inversions += (left-i);
            } else {
                arr[k++] = lptr[i++];
            }
        }
        while(i < left) arr[k++] = lptr[i++];
    }
    return inversions;
}