Counting Triangles

• Time : $O(nlogn + n^2)$
• Space : $O(1)$

int Solution::nTriang(vector<int> &A) {
    sort(begin(A), end(A));
    long long res = 0, mod = 1e9 + 7;
    for(int i = A.size() - 1; i >= 2; i--) {
        int l = 0, r = i - 1;
        while(l < r) {
            int sum = A[l] + A[r];
            if(sum > A[i]) {
                res = (res + (r - l) % mod) % mod;
                r--;
            } else l++;
        }
    }
    return res;
}