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:
\n\nx most frequent elements. If two elements have the same number of occurrences, the element with the bigger value is considered more frequent.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].
\n
Example 1:
\n\nInput: nums = [1,1,2,2,3,4,2,3], k = 6, x = 2
\n\nOutput: [6,10,12]
\n\nExplanation:
\n\n[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.[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.[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:
\n\nInput: nums = [3,8,7,8,7,5], k = 2, x = 2
\n\nOutput: [11,15,15,15,12]
\n\nExplanation:
\n\nSince k == x, answer[i] is equal to the sum of the subarray nums[i..i + k - 1].
\n
Constraints:
\n\nnums.length == n1 <= n <= 1051 <= nums[i] <= 1091 <= x <= k <= nums.lengthx frequent elements, and the second will contain all other elements.",
"Update the two sets whenever you slide the window, and maintain a sum of the elements in the set with x elements"
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