I think I've got it in O(n log n) so it is enough for n <= 2e5. Create prefix array in such a way that prefix[i] denotes (number of x's in prefix i — other numbers in prefix i). For exemplary array it will be 1 2 1 2 3. You create it as below:

for (int i = 1; i <= n; i++) {
        cin >> input;
        prefix[i] = prefix[i - 1];
        if (input == x)
            prefix[i]++;
        else
            prefix[i]--;
}

This is obviously O(n). Now let's go from the end of the array and for each i, count all the subarrays which start from i+1 and are proper. for(int i = n-1; i >= 0; i--) We can do it in O(log n). Number of subarrays that should be counted is equal to the number of such m: i < m <= n && prefix[m] — prefix[i] > 0. (it means that the prefix 'grows' so there are more x's than other numbers in the subarray). The results will be as below:

i = 4, count = 1
i = 3, count = 2
i = 2, count = 1
i = 1, count = 3
i = 0, count = 5
============== 12

How to know how many m's meet this condition for each i? We can keep them in tree. Leaves are all possible values of prefix[m] so there are n+1 of them. Every time we pass the prefix, we add it to the tree in O(log n). It guarantees that i < m <= n. When we want to count m's which meet the second condition, we can do it in O(log n), too. It's just a range query.