Sharing an answer code of mine about MaxProductOfThree problem of Codility lesson 6.

# Lesson 6: MaxProductOfThree

A non-empty zero-indexed array A consisting of N integers is given. The product of triplet (P, Q, R) equates to A[P] * A[Q] * A[R] (0 ≤ P < Q < R < N).

For example, array A such that:

\[A[0] = -3\\ A[1] = 1\\ A[2] = 2\\ A[3] = -2\\ A[4] = 5\\ A[5] = 6\\\]contains the following example triplets:

- (0, 1, 2), product is −3 * 1 * 2 = −6
- (1, 2, 4), product is 1 * 2 * 5 = 10
- (2, 4, 5), product is 2 * 5 * 6 = 60

Your goal is to find the maximal product of any triplet.

Write a function:

```
def solution(A)
```

that, given a non-empty zero-indexed array A, returns the value of the maximal product of any triplet.

For example, given array A such that:

\[A[0] = -3\\ A[1] = 1\\ A[2] = 2\\ A[3] = -2\\ A[4] = 5\\ A[5] = 6\\\]the function should return 60, as the product of triplet (2, 4, 5) is maximal.

Assume that:

- N is an integer within the range [3..100,000];
- each element of array A is an integer within the range [−1,000..1,000].

Complexity:

- expected worst-case time complexity is O(N*log(N));
- expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).

# Answer Code in Python 3

- Time complexity: \(O(N\log N)\)

```
# Time complexity: O(N*log(N))
# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")
import math
def solution(A):
# write your code in Python 3.6
max = -math.inf
A.sort()
for i in range(0, -4, -1):
if max < A[0 + i] * A[1 + i] * A[2 + i]:
max = A[0 + i] * A[1 + i] * A[2 + i]
return max
```