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

Python 100 / 100 (tested) on codility, with O(nlogn) time and O(n) space.

Here is @noisyboiler's python implementation of @Aryabhatta's method with comments and an example. Full credit to original authors, any errors / poor wording are entirely my fault.

from bisect import bisect_right

def number_of_disc_intersections(A):
    pairs = 0

    # create an array of tuples, each containing the start and end indices of a disk
    # some indices may be less than 0 or greater than len(A), this is fine!
    # sort the array by the first entry of each tuple: the disk start indices
    intervals = sorted( [(i-A[i], i+A[i]) for i in range(len(A))] )

    # create an array of starting indices using tuples in intervals
    starts = [i[0] for i in intervals]

    # for each disk in order of the *starting* position of the disk, not the centre
    for i in range(len(starts)):

        # find the end position of that disk from the array of tuples
        disk_end = intervals[i][1]

        # find the index of the rightmost value less than or equal to the interval-end
        # this finds the number of disks that have started before disk i ends
        count = bisect_right(starts, disk_end )

        # subtract current position to exclude previous matches
        # this bit seemed 'magic' to me, so I think of it like this...
        # for disk i, i disks that start to the left have already been dealt with
        # subtract i from count to prevent double counting
        # subtract one more to prevent counting the disk itsself
        count -= (i+1)
        pairs += count
        if pairs > 10000000:
            return -1
    return pairs

Worked example: given [3, 0, 1, 6] the disk radii would look like this:

disk0  -------         start= -3, end= 3
disk1      .           start=  1, end= 1
disk2      ---         start=  1, end= 3
disk3  -------------   start= -3, end= 9
index  3210123456789   (digits left of zero are -ve)

intervals = [(-3, 3), (-3, 9), (1, 1), (1,3)]
starts    = [-3, -3, 1, 1]

the loop order will be: disk0, disk3, disk1, disk2

0th loop: 
    by the end of disk0, 4 disks have started 
    one of which is disk0 itself 
    none of which could have already been counted
    so add 3
1st loop: 
    by the end of disk3, 4 disks have started 
    one of which is disk3 itself
    one of which has already started to the left so is either counted OR would not overlap
    so add 2
2nd loop: 
    by the end of disk1, 4 disks have started 
    one of which is disk1 itself
    two of which have already started to the left so are either counted OR would not overlap
    so add 1
3rd loop: 
    by the end of disk2, 4 disks have started
    one of which is disk2 itself
    two of which have already started to the left so are either counted OR would not overlap
    so add 0

pairs = 6
to check: these are (0,1), (0,2), (0,2), (1,2), (1,3), (2,3),