Maximize A[P] * A[Q] * A[R] for any triplet (P, Q, R).
- Difficulty Level: Easy
- Question URL: https://app.codility.com/programmers/lessons/6-sorting/max_product_of_three/
- Time Complexity:
Question:
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].
Solution:
Solution to Codility's Maximize Product Of Three problem which is from the Codility Lesson 6: Sorting and, is solved in Java 8 with 100% performance and correctness scores. The goal here is to maximize a[p] * a[q] * a[r] for any triplet (p, q, r). You can find the question of this MaxProductOfThree problem in the Codility website.
package Codility.Lesson6;
import java.util.*;
public class MaxProductOfThree {
public static void main(String[] args) {
// TODO Auto-generated method stub
int[] in2 = { 4, 5, 6 };
int result = solution(in2);
System.out.println(result);
}
public static int solution(int[] A) {
List aList = new ArrayList();
for(int i=0; i<A.length; i++) {
aList.add(A[i]);
}
Collections.sort(aList);
int product1, product2, product3, product4 = 0;
product1 = (int)aList.get(0) * (int)aList.get(1) * (int)aList.get(2); //first 3 elements
product2 = (int)aList.get(aList.size()-3) * (int)aList.get(aList.size()-2) * (int)aList.get(aList.size()-1); //last 3 elements
product3 = (int)aList.get(0) * (int)aList.get(1) * (int)aList.get(aList.size()-1); //first 2 and last element
product4 = (int)aList.get(0) * (int)aList.get(aList.size()-2) * (int)aList.get(aList.size()-1); //first and last 2 elements
int max1 = Math.max(product1, product2);
int max2 = Math.max(product3, product4);
return Math.max(max1, max2);
}
}