[Codility] Lesson 6 - MaxProductOfThree

Task discription

A non-empty 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 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.

Write an efficient algorithm for the following assumptions:

• 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].
  
  

Source code

#include <algorithm>

int solution(vector<int> &A) {
    sort(A.begin(), A.end(), greater<>());
    int endIdx = A.size()-1;
    int ans = A[0]*A[1]*A[2];
    // 음수*음수*양수가 최대값이 되는 경우 고려
    if(A[endIdx] < 0 && A[endIdx-1] < 0) {
        ans = max(ans, A[endIdx]*A[endIdx-1]*A[0]);
    }

    return ans;
}