Determine whether a triangle can be built from a given set of edges.
- Difficulty Level: Easy
- Question URL: https://app.codility.com/programmers/lessons/6-sorting/triangle/
- Time Complexity:
Question:
An array A consisting of N integers is given. A triplet (P, Q, R) is triangular if 0 ≤ P < Q < R < N and:
- A[P] + A[Q] > A[R],
- A[Q] + A[R] > A[P],
- A[R] + A[P] > A[Q].
For example, consider array A such that:
A[0] = 10 A[1] = 2 A[2] = 5 A[3] = 1 A[4] = 8 A[5] = 20
Triplet (0, 2, 4) is triangular.
Write a function:
class Solution { public int solution(int[] A); }
that, given an array A consisting of N integers, returns 1 if there exists a triangular triplet for this array and returns 0 otherwise.
For example, given array A such that:
A[0] = 10 A[1] = 2 A[2] = 5 A[3] = 1 A[4] = 8 A[5] = 20
the function should return 1, as explained above. Given array A such that:
A[0] = 10 A[1] = 50 A[2] = 5 A[3] = 1
the function should return 0.
Write an efficient algorithm for the following assumptions:
- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].
Solution:
Solution to Codility's Triangle 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 determine whether a triangle can be built from a given set of edges. You can find the question of this Triangle problem in the Codility website.
package Codility.Lesson6;
import java.util.*;
public class Triangle {
public static void main(String[] args) {
// TODO Auto-generated method stub
int[] in2 = {8, 4, 5, 6,10,15,25 };
int result = solution(in2);
System.out.println(result);
}
public static int solution(int[] A) {
// write your code in Java SE 8
if (A.length < 3)
return 0;
Arrays.sort(A);
// made long array because each int element can be as high as
// Integer MAX_VALUE so when add them can overflow int
long[] aOrdered = new long[A.length];
int index = 0;
for (Integer i : A) {
aOrdered[index++] = i;
}
// start from the end (largest)
// if previous 2 elements have sum > current element, found a triangle
for (int i = aOrdered.length - 1; i >= 2; i--) {
if (aOrdered[i - 1] + aOrdered[i - 2] > aOrdered[i]) {
return 1;
}
}
return 0;
}
}