The Question:
You are given a 0-indexed integer array nums.
The distinct count of a subarray of nums is defined as:
Let nums[i..j] be a subarray of nums consisting of all the indices from i to j such that 0 <= i <= j < nums.length. Then the number of distinct values in nums[i..j]is called the distinct count of nums[i..j].
Return the sum of the squares of distinct counts of all subarrays of nums.
A subarray is a contiguous non-empty sequence of elements within an array.
This is the solution online since I wasn’t able to solve it myself but I don’t understand how it takes all the subarrays into consideration:
Solution:

    int sumCounts(vector<int>& nums) {
        int n = nums.size();
        int result = 0;
        for (int i = 0; i < n; i++) {
            unordered_map<int, int> hash;
            for (int j = i; j < n; j++) {
                hash[nums[j]]++;
                result += (hash.size() * hash.size());
            }
        }
        return result;
    }

for example: [1,2,1]
it considers:[1,2,1]
then: [2,1]
and then [1]
but what about [2] and [1]?
I’m very confused where I am going wrong. Kindly help me understand it.