# Codility ‘NumberOfDiscIntersections’ 2010 Beta Solution

##### 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.

##### Link

##### Complexity:

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

expected worst-case space complexity is O(N)

##### Execution:

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.

##### Solution:

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 else: active_circles -= 1 if intersections > 10E6: return -1 return intersections

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