codility lesson 4 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.

import java.util.*;
class Solution {
    public int solution(int[] A) {
        // write your code in Java 
        if(A == null || A.length <3)
            return 0;
        Arrays.sort(A);
        for(int i = 0; i < A.length - 2; i++)
            if(A[i] >= 1073741824 || A[i+1] >= 1073741824) { // in case of overflow
                // double a = A[i] + A[i+1]; // this does not work
                double a = A[i]; // notice that if use A[i] + A[i+1] then they will get -2 first before give a the                                     correct value
                double b = A[i+1];
                //System.out.println(a+b);
                if(a + b > A[i+2])
                    return 1;
            }else{
                if(A[i] > 0 && A[i] + A[i+1] > A[i+2])
                    return 1;
            }
        return 0;
    }
}