Codility 'NumberOfDiscIntersections' 2010 Beta Solution

Martin Kysel · April 27, 2015

Short Problem Definition:

Compute intersections between sequence of discs.

Given an array A of N integers, we draw N discs in a 2D plane such that the I-th disc is centered on (0,I) and has a radius of A[I]. We say that the J-th disc and K-th disc intersect if J ≠ K and J-th and K-th discs have at least one common point.

Number of Disc Intersections


expected worst-case time complexity is O(N\*log(N))

expected worst-case space complexity is O(N)


In my mind this problem is very similar to the Codility Fish. You keep count of how many not-yet-ended disks there are. Every beginning has to cross all not-yet-ended disks. The best explanation on the web was written by Luca. His solution is much better than my C-style double-checked while loop. I therefore created a mixture of both :)

  • Keep in mind that any pair of circles can intersect only once.
  • If 1+ beginnings have the same point as 1+ endings. You first have to process the beginnings. The old and new ones have per definition a meeting point.

def solution(A):
    circle_endpoints = []
    for i, a in enumerate(A):
        circle_endpoints += [(i-a, True), (i+a, False)]

    circle_endpoints.sort(key=lambda x: (x[0], not x[1]))

    intersections, active_circles = 0, 0

    for _, is_beginning in circle_endpoints:
        if is_beginning:
            intersections += active_circles
            active_circles += 1
            active_circles -= 1
        if intersections > 10E6:
            return -1

    return intersections

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