CountTriangles

An array A consisting of N integers is given. A triplet (P, Q, R) is triangular if it is possible to build a triangle with sides of lengths A[P], A[Q] and A[R]. In other words, 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] = 12

There are four triangular triplets that can be constructed from elements of this array, namely (0, 2, 4), (0, 2, 5), (0, 4, 5), and (2, 4, 5).

Write a function:

    int solution(vector<int> &A);

that, given an array A consisting of N integers, returns the number of triangular triplets in this array.

For example, given array A such that:
  A[0] = 10    A[1] = 2    A[2] = 5
  A[3] = 1     A[4] = 8    A[5] = 12

the function should return 4, as explained above.

Write an efficient algorithm for the following assumptions:

        N is an integer within the range [0..1,000];
        each element of array A is an integer within the range [1..1,000,000,000].

#include <algorithm>

int solution(vector<int> &A) {
    // write your code in C++11 (g++ 4.8.2)
    int s = int(A.size());
    
    if (s < 3) return 0;
    
    sort(A.begin(), A.end());
    
    if (A[0] + A[1] > A[s - 1]) {
        return s * (s - 1) * (s -2) / 6;
    }
    
    int num = 0;
    int start = 0;
    int end = 0;
    for (int i = 0; i < s -2; i++) {
        for (int j = i + 1; j < s - 1; j++) {
            for (int k = j + 1; k < s; k++) {
                if (A[i] + A[j] > A[k]) {
                    num++;
                } else {
                    break;
                }
            }
        }
    }
    
    return num;
}