Codility - Lesson 15 Caterpillar Method - 2. CountTriangles

Source Link:
https://app.codility.com/programmers/lessons/15-caterpillar_method/count_triangles/

Question:

A zero-indexed 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:

class Solution { public int solution(int[] A); }

that, given a zero-indexed 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.

Assume that:

• 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].

Complexity:

• expected worst-case time complexity is O(N2);
• expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

My Solution:
Notes:
1. // important: sort the edges, so that we just need to check if "1st edge + 2nd edge > 3rd edge"
Arrays.sort(A);
2. // Using "Caterpillar method", so we can have O(n^2), not O(n^3)
// key point of "Caterpillar method"
if(rightEnd < A.length && A[i] + A[leftEnd] > A[rightEnd]){
rightEnd++; // increase the Caterpillar
}
else{
numTriangle = numTriangle + (rightEnd - leftEnd - 1);
leftEnd++; // decrease the Caterpillar
}
3.
return numTriangle;