Find X-Sum of All K-Long Subarrays I

You are given an array `nums` of `n` integers and two integers `k` and `x`. The **x-sum** of an array is calculated by the following procedure: - Count the occurrences of all elements in the array. - Keep only the occurrences of the top `x` most frequent elements. If two elements have the same number of occurrences, the element with the **bigger** value is considered more frequent. - Calculate the sum of the resulting array. **Note** that if an array has less than `x` distinct elements, its **x-sum** is the sum of the array. Return an integer array `answer` of length `n - k + 1` where `answer[i]` is the **x-sum** of the subarray `nums[i..i + k - 1]`.   **Example 1:** **Input:** nums = [1,1,2,2,3,4,2,3], k = 6, x = 2 **Output:** [6,10,12] **Explanation:** - For subarray `[1, 1, 2, 2, 3, 4]`, only elements 1 and 2 will be kept in the resulting array. Hence, `answer[0] = 1 + 1 + 2 + 2`. - For subarray `[1, 2, 2, 3, 4, 2]`, only elements 2 and 4 will be kept in the resulting array. Hence, `answer[1] = 2 + 2 + 2 + 4`. Note that 4 is kept in the array since it is bigger than 3 and 1 which occur the same number of times. - For subarray `[2, 2, 3, 4, 2, 3]`, only elements 2 and 3 are kept in the resulting array. Hence, `answer[2] = 2 + 2 + 2 + 3 + 3`. **Example 2:** **Input:** nums = [3,8,7,8,7,5], k = 2, x = 2 **Output:** [11,15,15,15,12] **Explanation:** Since `k == x`, `answer[i]` is equal to the sum of the subarray `nums[i..i + k - 1]`.   **Constraints:** - `1 <= n == nums.length <= 50` - `1 <= nums[i] <= 50` - `1 <= x <= k <= nums.length`