Test results - Codility

Tasks Details

easy

1. MaxProductOfThree

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

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 8

Time spent on task 1 minutes

Notes

not defined yet

Task timeline

Code: 03:16:26 UTC, java, verify, result: Passed

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

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

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
import java.util.Arrays;
class Solution {
	public int solution(int[] A) {
		Arrays.sort(A);

		int N = A.length;
		int firstProduct = A[0] * A[1];
		int secondProduct = A[N - 3] * A[N - 2];
		int lastProduct = A[N -1];

		if(lastProduct < 0) return Math.min(firstProduct, secondProduct) * A[N -1];
		else return Math.max(firstProduct, secondProduct) * A[N -1];
	}
}

Analysis

▶

example
example test

✔

Code: 03:16:33 UTC, java, verify, result: Passed

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

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

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
import java.util.Arrays;
class Solution {
	public int solution(int[] A) {
		Arrays.sort(A);

		int N = A.length;
		int firstProduct = A[0] * A[1];
		int secondProduct = A[N - 3] * A[N - 2];
		int lastProduct = A[N -1];

		if(lastProduct < 0) return Math.min(firstProduct, secondProduct) * A[N -1];
		else return Math.max(firstProduct, secondProduct) * A[N -1];
	}
}

Analysis

▶

example
example test

✔

Code: 03:16:37 UTC, java, final, score: 100

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

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

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
import java.util.Arrays;
class Solution {
	public int solution(int[] A) {
		Arrays.sort(A);

		int N = A.length;
		int firstProduct = A[0] * A[1];
		int secondProduct = A[N - 3] * A[N - 2];
		int lastProduct = A[N -1];

		if(lastProduct < 0) return Math.min(firstProduct, secondProduct) * A[N -1];
		else return Math.max(firstProduct, secondProduct) * A[N -1];
	}
}

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

O(N * log(N))

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example
example test

✔

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one_triple
three elements

✔

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simple1
simple tests

✔

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simple2
simple tests

✔

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small_random
random small, length = 100

✔

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medium_range
-1000, -999, ... 1000, length = ~1,000

✔

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medium_random
random medium, length = ~10,000

✔

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large_random
random large, length = ~100,000

✔

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large_range
2000 * (-10..10) + [-1000, 500, -1]

✔

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extreme_large
(-2, .., -2, 1, .., 1) and (MAX_INT)..(MAX_INT), length = ~100,000

✔