Codility ‘CountTriangles’ Solution

Short Problem Definition:

Count the number of triangles that can be built from a given set of edges.

Link

CountTriangles

Complexity:

expected worst-case time complexity is O(N2);

expected worst-case space complexity is O(1)

Execution:

Apply the caterpillar method. We know that in a sorted array every position between Q and R will be bigger than Q and therefore P+Q will be bigger than R. I therefore either increment Q if P+Q is not larger than R or increment R as far as possible.

Solution:
def solution(A):
    A.sort()
    print A
    triangle_cnt = 0
    
    for P_idx in xrange(0, len(A)-2):
        Q_idx = P_idx + 1
        R_idx = P_idx + 2
        while (R_idx < len(A)):
            if A[P_idx] + A[Q_idx] > A[R_idx]:
                triangle_cnt += R_idx - Q_idx
                R_idx += 1
            elif Q_idx < R_idx -1:
                    Q_idx += 1
            else:
                R_idx += 1
                Q_idx += 1
                
    return triangle_cnt

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