Algorithm to calculate number of intersecting discs

Question

Given an array A of N integers we draw N discs in a 2D plane, such that i-th disc has center in (0,i) and a radius

Answer 1

Here's a two-pass C++ solution that doesn't require any libraries, binary searching, sorting, etc.

int solution(vector &A) {
    #define countmax 10000000
    int count = 0;
    // init lower edge array
    vector E(A.size());
    for (int i = 0; i < (int) E.size(); i++)
        E[i] = 0;
    // first pass
    // count all lower numbered discs inside this one
    // mark lower edge of each disc
    for (int i = 0; i < (int) A.size(); i++)
    {
        // if disc overlaps zero
        if (i - A[i] <= 0)
            count += i;
        // doesn't overlap zero
        else {   
            count += A[i];
            E[i - A[i]]++;
        }
        if (count > countmax)
            return -1;
    }
    // second pass
    // count higher numbered discs with edge inside this one
    for (int i = 0; i < (int) A.size(); i++)
    {
        // loop up inside this disc until top of vector
        int jend = ((int) E.size() < (long long) i + A[i] + 1 ? 
                    (int) E.size() : i + A[i] + 1);
        // count all discs with edge inside this disc
        // note: if higher disc is so big that edge is at or below 
        // this disc center, would count intersection in first pass
        for (int j = i + 1; j < jend; j++)
            count += E[j];
        if (count > countmax)
            return -1;
    }
    return count;
}