Finding pairs with product greater than sum

Question

Given as input, a sorted array of floats, I need to find the total number of pairs (i,j) such as A[i]*A[j]>=A[i]+A[j] for each i < j

Answer 1

Here's a O(n) algorithm that solves the problem when the array's elements are positive.

When the elements are positive, we can say that:

• If A[i]*A[j] >= A[i]+A[j] when j>i then A[k]*A[j] >= A[k]+A[j] for any k that satisfies k>i (because the array is sorted).

• If A[i]*A[j] < A[i]+A[j] when j>i then A[i]*A[k] < A[i]+A[k] for any k that satisfies k.

(these facts don't hold when both numbers are fractions, but then the condition won't be satisfied anyway)

Thus we can perform the following algorithm:

int findNumOfPairs(float A[])
{    
    start = 0;
    end = A.length - 1;
    numOfPairs = 0;

    while (start != end)
    {
        if (A[start]*A[end] >= A[start]+A[end])
        {
            numOfPairs += end - start;
            end--;
        }
        else
        {
            start++;
        }
    }

    return numOfPairs;
}