Nah, if you do like that, your code would absolutely get TLE (or in this case is WA).

So here is the basic ideas: you are given an array $$$a$$$ with $$$n$$$ elements then the element $$$a_i$$$ will be contained by exactly $$$(n-1-i) \times i+n-1$$$ segments and all the elements between $$$a_i$$$ and $$$a_{i+1}$$$ will be contained by exactly $$$(n-i) \times i$$$ segments. Also, in this case you only need to count how many points that contained by exactly $$$k$$$ segments so you can use map for the efficiency

For example, The first element $$$a[0]$$$ will be contained by exactly $$$n-1$$$ segments so m[n-1]++. All the elements in $$$a[0]$$$ and $$$a[1]$$$ will be contained by exactly $$$(n-1)*1$$$ segments so m[(n-1)*1] += (a[1] - a[0] - 1). Then the element $$$a[1]$$$ will be contained by exactly $$$(n-2)*1 + n-1$$$ so m[(n-2)*1 + n-1]++, etc.

m[n-1]++;
for (int i = 1; i < n; ++i) {
    m[(n-i)*i] += (x[i] - x[i-1] - 1);
    m[(n-1-i)*i+n-1]++;
}

And since there are only $$$10^5$$$ elements so the size of the map $$$m$$$ can only be $$$\leq 2*n$$$. Therefore, if $$$k$$$ is exist in the map then the answer is $$$m[k]$$$ otherwise, $$$0$$$

You can refer to my code if needed (it is cleaned and easy to visualize) 283198814