maxproductofthree
Description
A non-empty zero-indexed 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:
Write a function:
class Solution { public int solution(int[] A); }
that, given a non-empty zero-indexed 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.
Assume that:
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
Write a function:
class Solution { public int solution(int[] A); }
that, given a non-empty zero-indexed 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.
Assume that:
- 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].
- expected worst-case time complexity is O(N*log(N));
- expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).
Codes (100%)
Pitfall: two negative numbers multiplied together are positive.
import java.util.*;
class Solution {
public int solution(int[] A) {
// write your code in Java SE 8
Arrays.sort(A);
return Math.max(A[A.length-1]*A[A.length-2]*A[A.length-3], A[0]*A[1]*A[A.length-1]);
}
}
Pitfall: two negative numbers multiplied together are positive.
import java.util.*;
class Solution {
public int solution(int[] A) {
// write your code in Java SE 8
Arrays.sort(A);
return Math.max(A[A.length-1]*A[A.length-2]*A[A.length-3], A[0]*A[1]*A[A.length-1]);
}
}