Tasks Details
easy
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] = 6contains 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] = 6the 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].
Copyright 2009–2024 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used Java 8
Time spent on task 2 minutes
Notes
not defined yet
Task timeline
Code: 09:53:09 UTC,
java,
autosave
// 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");
class Solution {
public int solution(int[] A) {
// write your code in Java SE 8
Arrays.sort(A);
if(A[A.length-1] < 0){
return A[A.length-1] * A[A.length-2] * A[A.length-3];
}
int minusValue = A[0] * A[1];
int plusValue = A[A.length-3] * A[A.length-2];
if (plusValue < minusValue) {
return A[A.length-1] * minusValue;
}else{
return A[A.length-1] * plusValue;
}
}
}
Code: 09:53:13 UTC,
java,
verify,
result: Passed
// you can also use imports, for example:
// import java.util.*;
import java.util.Arrays;
// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
class Solution {
public int solution(int[] A) {
// write your code in Java SE 8
Arrays.sort(A);
if(A[A.length-1] < 0){
return A[A.length-1] * A[A.length-2] * A[A.length-3];
}
int minusValue = A[0] * A[1];
int plusValue = A[A.length-3] * A[A.length-2];
if (plusValue < minusValue) {
return A[A.length-1] * minusValue;
}else{
return A[A.length-1] * plusValue;
}
}
}
Analysis
Code: 09:53:33 UTC,
java,
autosave
// you can also use imports, for example:
// import java.util.*;
import java.util.Arrays;
// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
class Solution {
public int solution(int[] A) {
// write your code in Java SE 8
Arrays.sort(A);
if(A[A.length-1] < 0){
return A[A.length-1] * A[A.length-2] * A[A.length-3];
}else{
int minusValue = A[0] * A[1];
int plusValue = A[A.length-3] * A[A.length-2];
if (plusValue < minusValue) {
return A[A.length-1] * minusValue;
}else{
return A[A.length-1] * plusValue;
}
}
}
}
Code: 09:53:58 UTC,
java,
verify,
result: Passed
// you can also use imports, for example:
// import java.util.*;
import java.util.Arrays;
// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
class Solution {
public int solution(int[] A) {
// write your code in Java SE 8
Arrays.sort(A);
if(A[A.length-1] < 0){
return A[A.length-1] * A[A.length-2] * A[A.length-3];
}
int minusValue = A[0] * A[1];
int plusValue = A[A.length-3] * A[A.length-2];
if (plusValue < minusValue) {
return A[A.length-1] * minusValue;
}else{
return A[A.length-1] * plusValue;
}
}
}
Analysis
Code: 09:54:04 UTC,
java,
verify,
result: Passed
// you can also use imports, for example:
// import java.util.*;
import java.util.Arrays;
// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
class Solution {
public int solution(int[] A) {
// write your code in Java SE 8
Arrays.sort(A);
if(A[A.length-1] < 0){
return A[A.length-1] * A[A.length-2] * A[A.length-3];
}
int minusValue = A[0] * A[1];
int plusValue = A[A.length-3] * A[A.length-2];
if (plusValue < minusValue) {
return A[A.length-1] * minusValue;
}else{
return A[A.length-1] * plusValue;
}
}
}
Analysis
Code: 09:54:09 UTC,
java,
final,
score:
100
// you can also use imports, for example:
// import java.util.*;
import java.util.Arrays;
// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
class Solution {
public int solution(int[] A) {
// write your code in Java SE 8
Arrays.sort(A);
if(A[A.length-1] < 0){
return A[A.length-1] * A[A.length-2] * A[A.length-3];
}
int minusValue = A[0] * A[1];
int plusValue = A[A.length-3] * A[A.length-2];
if (plusValue < minusValue) {
return A[A.length-1] * minusValue;
}else{
return A[A.length-1] * plusValue;
}
}
}
Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N * log(N))
expand all
Correctness tests
1.
0.004 s
OK
2.
0.004 s
OK
3.
0.004 s
OK
1.
0.004 s
OK
2.
0.004 s
OK
3.
0.004 s
OK
4.
0.008 s
OK
1.
0.004 s
OK
2.
0.004 s
OK
3.
0.004 s
OK
1.
0.004 s
OK
expand all
Performance tests
1.
0.016 s
OK
1.
0.028 s
OK
1.
0.236 s
OK
1.
0.080 s
OK
1.
0.156 s
OK
2.
0.228 s
OK