Compute iterative sliding-window sum reduction

## Problem Given an integer array `nums` and an integer window size `k`, repeatedly apply the following reduction: - If `len(nums) > k`: replace `nums` with a new array `next` where - `next[i] = sum(nums[i : i + k])` for every valid consecutive window of length `k`. - Thus, `len(next) = len(nums) - k + 1`. - If `len(nums) <= k`: stop and return `sum(nums)`. Return the final sum. ### Example `nums = [1, 1, 1, 2, 1, 2]`, `k = 3` Round 1 (window size 3): - `[1,1,1] -> 3`, `[1,1,2] -> 4`, `[1,2,1] -> 4`, `[2,1,2] -> 5` - New array: `[3, 4, 4, 5]` Round 2 (window size 3): - `[3,4,4] -> 11`, `[4,4,5] -> 13` - New array: `[11, 13]` Stop because `len([11,13]) = 2 <= 3`, return `11 + 13 = 24`. ### Function Signature Implement a function like: - `int reduceSlidingWindowSum(int[] nums, int k)` ### Notes / Constraints (assume) - `nums.length >= 1` - `1 <= k` - Values can be negative; use a wide enough integer type if overflow is possible.