Codility 'MaxProductOfThree' Solution

Martin Kysel · August 6, 2014

Short Problem Definition:

Maximize A[P] * A[Q] * A[R] for any triplet (P, Q, R).



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

expected worst-case space complexity is O(1)


After sorting the largest product can be found as a combination of the last three elements. Additionally, two negative numbers add to a positive, so by multiplying the two largest negatives with the largest positive, we get another candidate. If all numbers are negative, the three largest (closest to 0) still get the largest element!


def solution(A):
    if len(A) < 3:
        raise Exception("Invalid input")
    return max(A[0] * A[1] * A[-1], A[-1] * A[-2] * A[-3])

A better solution has been suggested in the commends by Mindaugas Tvaronavičius. Here the solution in O(N) time using two heaps.

def betterSolution(A):
    if len(A) < 3:
        raise Exception("Invalid input")
    minH = []
    maxH = []
    for val in A:
        if len(minH) < 2:
            heapq.heappush(minH, -val)
            heapq.heappushpop(minH, -val)
        if len(maxH) < 3:
            heapq.heappush(maxH, val)
            heapq.heappushpop(maxH, val)
    max_val = heapq.heappop(maxH) * heapq.heappop(maxH)
    top_ele = heapq.heappop(maxH)
    max_val *= top_ele
    min_val = -heapq.heappop(minH) * -heapq.heappop(minH) * top_ele
    return max(max_val, min_val)

Twitter, Facebook