Given an array of integers, nums, determine the number of triplets of distinct indices (i, j, k) that can be selected from the array such that the values at those indices can form the sides of a valid triangle.

Return this count as the result.

Note: The triplet of indices is treated as an unordered combination, not a permutation. For example, indices (0, 1, 2) and (2, 1, 0) represent the same triplet and should be counted once.

Constraints:

• $1 <=$ nums.length $<= 1000$

• $0 <=$ nums[i] $<= 1000$

The solution uses the sort and search pattern to count the number of valid triangles formed from a given set of numbers. The triangle inequality rule states that a triangle is valid if the sum of the two smaller sides is greater than the largest. Sorting the array in ascending order simplifies the problem, as the largest side in any triplet (group of three) is always at the end. After sorting and considering the triangle inequality rule, we iterate backward from the end of the array to perform the search for valid combinations. While iterating, we identify valid combinations by fixing the current element as the largest side and exploring potential pairs for the other two smaller sides.

Now, let’s look at the solution steps below:

1. We begin by sorting the array nums. Sorting is important because it ensures that for any triplet (nums[left], nums[right], nums[i]) where i > right > left, nums[i] is the largest side. This allows us to only check if nums[left] + nums[right] > nums[i].

2. We initialize a variable, count, to $0$ for tracking the number of valid triangles.

3. We iterate backward from the end of nums, treating ...