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Lesson06(Sorting)-MaxProductOfThree.cpp
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Lesson06(Sorting)-MaxProductOfThree.cpp
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// 2. MaxProductOfThree.
/**
* 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.
*
* 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].
*/
#include <vector>
#include <algorithm>
int maxProductOfThree(std::vector<int>& A)
{
int n = (int)A.size();
if (n < 3) {
return 0;
}
// Sort the input vector in ascending order.
std::sort(std::begin(A), std::end(A));
// The maximum product of the 3 elements in a sorted vector can take on one of 4 values:
// 1. Multiplying the first 3 elements.
// 2. Multiplying the last 3 elements (which may signify the highest positive values).
// 3. Multiplying the first element with the last 2 elements.
// 4. Multiplying the first 2 elements with the last element
// (accounting for the potential scenario of 2 first large negative values becoming positive when multiplied
// and also multiplied with the last positive value).
int tripletProduct1 = A[0] * A[1] * A[2],
tripletProduct2 = A[n - 1] * A[n - 2] * A[n - 3],
tripletProduct3 = A[0] * A[n - 1] * A[n - 2],
tripletProduct4 = A[0] * A[1] * A[n - 1];
return std::max(std::max(tripletProduct1, tripletProduct2),
std::max(tripletProduct3, tripletProduct4));
}