Problem Description

Given an integer array nums and an integer window size k, repeatedly apply the following reduction process:

1. Iteration Condition: If the current length of nums is greater than k, proceed to the next step. Otherwise, stop the process and return the sum of all elements in the current nums array.
2. Reduction Step: Create a new array, next. For each valid consecutive window of length k in the current nums array, calculate its sum and store it as an element in next.
   • Specifically, next[i] will be the sum of nums[i] through nums[i + k - 1].
   • The length of the next array will be len(nums) - k + 1.
3. Update: Replace nums with the newly computed next array and repeat the process from step 1.

The goal is to return the final sum when the process terminates.

Example

nums = [1, 1, 1, 2, 1, 2], k = 3

Round 1 (window size 3):

• Windows and their sums: [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):

• Windows and their sums: [3,4,4] -> 11, [4,4,5] -> 13
• New array: [11, 13]

Termination: The length of [11, 13] is 2, which is less than or equal to k=3. The process stops. Final Result: Return 11 + 13 = 24.

Function Signature

int reduceSlidingWindowSum(int[] nums, int k)

Constraints

• nums.length >= 1
• 1 <= k
• Array elements can be negative. Implementations should consider using a data type wide enough to prevent integer overflow for sums.