Distinct Values on Maximum-Sum Frames

For this exercise, use the callable contract below.
Given an integer matrix and frameSize, consider every contiguous frameSize × frameSize submatrix. Its border contains each cell on its top row, bottom row, left column, or right column exactly once.
Find the maximum border sum. Among every frame tied for that maximum, collect all integer values that occur on a winning border. Return the sum of those distinct values.

Function
sumDistinctWinningBorderValues(matrix: int[][], frameSize: int) → long

Examples
Example 1
matrix = [[1,2,3],[4,5,6],[7,8,9]]
frameSize = 2
return = 28
The bottom-right 2-by-2 frame has the unique maximum border sum, and 5+6+8+9=28.
Example 2
matrix = [[1,1,1],[1,0,1],[1,1,1]]
frameSize = 2
return = 1
All four frames tie; the union of their border values is {0,1}.

Constraints
1 <= rows, columns <= 200.
1 <= frameSize <= min(rows, columns).
Matrix values fit in a 32-bit signed integer; the result fits in a 64-bit signed integer.

The trick is computing each border sum in O(1). Build row prefix sums and column prefix sums. For a frame at (r, c) with size k, border = top row segment + bottom row segment + left column segment + right column segment, minus the four corners you counted twice. Watch k=1, where the border is a single cell, and k=2, where the overlap handling changes. Easiest fix: special-case k=1, and otherwise subtract corners carefully, or sum the left and right columns excluding the top and bottom rows. Pass one finds the max sum. Pass two walks the tied frames and adds border values into a hash set, then sums the set into a 64-bit long. Pitfall: overflow from using int, and re-walking borders for non-tied frames. Worst case is about 200x200 frames times 4k cells for the set pass, which is fine. If you blank mid-assessment, StealthCoder is the hedge, but know the corner subtraction cold.

Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.