values of length n and two integers width and top.
For any segment of the array, define its **frequency sum** as follows:
1. Count how many times each value occurs.
2. Rank the distinct values by frequency; on a tie, the larger value ranks higher.
3. Keep only the top highest-ranked values (all of their occurrences) and add up the kept elements.
If the segment has fewer than top distinct values, its frequency sum is simply the sum of the segment.
Return an array of n - width + 1 numbers whose i-th entry is the frequency sum of values[i..i + width - 1].values=[4, 4, 1, 7, 1, 4], width=4, top=1 → [8, 2, 2]
Windows: [4,4,1,7] — 4 is most frequent, sum 8; [4,1,7,1] — 1 is most frequent, sum 2; [1,7,1,4] — again 1, sum 2.
values=[2, 9, 2, 9, 5], width=5, top=2 → [22]
There is a single window. 2 and 9 occur twice each, 5 once. Keeping two groups gives 2 + 2 + 9 + 9 = 22.
values=[3, 6, 8], width=2, top=1 → [6, 8]
In each window every number occurs once, so the larger one wins: 6, then 8.
n = values.length ≤ 50values[i] ≤ 50top ≤ width ≤ nvalues=[4, 4, 1, 7, 1, 4], width=4, top=1
[8, 2, 2]
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