Codility - Lesson 6 Sorting - 3. MaxProductOfThree

Source Link:
https://codility.com/programmers/lessons/6-sorting/max_product_of_three/

Question:

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:

class Solution { public int solution(int[] 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).

Elements of input arrays can be modified.

My Solution:
Notes:
1. main idea:
// max_1 = positive * positive * positive
// max_2 = negative * negative * positive
// max = Math.max(max_1, max_1)
// just need to sort the integer array   
2. sort the array
Arrays.sort(A);
3. max_1 = 1st biggest * 2nd biggest * 3rd biggest
int max_1 = A[A.length-1] * A[A.length-2] * A[A.length-3];
4. max_2 = 1st smallest * 2nd smallest * 1st biggest
int max_2 = A[0] * A[1] * A[A.length-1];
5. take the maximum
int max = Math.max(max_1, max_2);