Codility - Lesson 6 Sorting - 2. Triangle

Source Link:
https://codility.com/programmers/lessons/6-sorting/triangle/

Question:

A zero-indexed 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 a zero-indexed 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.

Assume that:

• 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].

Complexity:

• expected worst-case time complexity is O(N*log(N));
• expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

My Solution:
Notes:
1. main idea: for any combination (A[i-2], A[i-1], A[i]) 
// we just need to check if A[i-2] + A[i-1] > A[i] (important)
// A[i-2] + A[i-1] is the max possible combination (needed to check) 
2. Using "Arrays.sort(int[])"
Arrays.sort(A);
3. start from i=2
for(int i=2; i< A.length; i++){
if((long)A[i-2] + (long)A[i-1] > (long)A[i]) // note: using "long" for overflow cases
return 1; 
// note: we just need one combination
}