Test results - Codility

Tasks Details

easy

1. MaxProductOfThree

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

Task Score

100%

Correctness

100%

Performance

100%

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

Programming language used Java 21

Time spent on task 11 minutes

Notes

not defined yet

Code: 07:49:46 UTC, java, verify, result: Failed

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// you can also use imports, for example:
// import java.util.*;

// you can use System.out.println for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
    public int solution(int[] A) {
        int[] maxes = new int[3]; // maxes[0] <= maxes[1] <= maxes[2]
        int[] mins = new int[2];  // mins [0] <= mins [1]
        boolean foundZero = false;

        // O(n)        
        for( int a : A )
        {
            updateMaxes( a, maxes );
            updateMins( a, mins );
        }
 
        System.out.println(Arrays.toString(maxes));
        System.out.println(Arrays.toString(mins));
        
        int obvious = maxes[0] * maxes[1] * maxes[2];
        int twoBigNegs = mins[0] * mins[1] * maxes[2];
        return Math.max( obvious, twoBigNegs );
    }
    
    private static void updateMaxes( int a, int[] maxes )
    {
        if( a >= maxes[2] ) {
            // Found new max, shift down
            maxes[0] = maxes[1];
            maxes[1] = maxes[2];
            maxes[2] = a;
        } else if( a >= maxes[1] ) {
            maxes[0] = maxes[1];
            maxes[1] = a;
        } else if( a > maxes[0] ) {
            maxes[0] = a;
        }
    }

    private static void updateMins( int a, int[] mins )
    {
        if( a <= mins[0] ) {
            // Found new min, shift down
            mins[1] = mins[0];
            mins[0] = a;
        } else if( a < mins[1] ) {
            mins[1] = a;
        }
    }
}