[GO] [100%] MaxProductOfThree (src & jUnit)

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MaxProductOfThree

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

Task description

Given a non-empty array A consisting of N integers. The product of triplets (P, Q, R) is equal to A [P] * A [Q] * A [R](0 ≤ P <Q <R <N).

For example, the following array A:

Example

   A [0] = -3
   A [1] = 1
   A [2] = 2
   A [3] = -2
   A [4] = 5
   A [5] = 6

The following triplet example is included.

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

Your goal is to find the maximum product of triplets.

Write a function:

class solution {public int solution(int [] A); }

Given a non-empty array A, returns the value of the maximum product of any triplet.

For example, given the following array A:

Example

   A [0] = -3
   A [1] = 1
   A [2] = 2
   A [3] = -2
   A [4] = 5
   A [5] = 6

The function must return 60 because the product of triplets (2, 4, 5) is the largest.

Write a ** efficient ** algorithm for the following assumptions:

• N is an integer in the range [3..100,000].
• Each element of array A is an integer in the range [-1,000..1,000].

solve

Program MaxProductOfThreeSolution.java

    public int solution(int[] A) {
        int max1 = -9999;
        int max2 = -9999;
        int max3 = -9999;
        int min2 = 9999;
        int min1 = 9999;

        for (int a : A) {
            if (a > max1) {
                max3 = max2;
                max2 = max1;
                max1 = a;
            } else if (a > max2) {
                max3 = max2;
                max2 = a;
            } else if (a > max3) {
                max3 = a;
            }

            if (a < min1) {
                min2 = min1;
                min1 = a;
            } else if (a < min2) {
                min2 = a;
            }
        }

        int maxProduct = max1 * max2 * max3;

        if (max1 <= 0 || min1 >= 0) {
            //All negative||All positive
            return maxProduct;
        } else {
            if (min2 < 0 ) {
                //At least two negatives
                int temp = max1 * min1 * min2;
                if (temp > maxProduct) return temp;
            }
        }

        return maxProduct;
    }

Detected time complexity:

O(N * log(N))

jUnit MaxProductOfThreeSolutionTest.java

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