Fix the window length in advance
Updated
k is given. Every window has length k. You do not discover length. You slide a block of k cells.
What to focus on
Length is always k = R-L+1. For n cells there are n-k+1 windows.
Worked examples
k=2 on four
a = [1, 2, 3, 4], k=2- windows [0,1], [1,2], [2,3]
- each has length 2
- count = 4-2+1 = 3
Result: 3 windows
k=2 on three
a = [4, 1, 9], k=2- [0,1] then [1,2]
- count = 2
Result: 2 windows
k bigger than n
a = [7], k=2- n=1 < 2
- no block of 2 cells
Result: 0 windows
Try it
Write windowCount(n, k). n is the array length. k is the fixed window length. Each window includes k cells, R-L+1 = k. Count is max(0, n-k+1). Return the count. On [1, 2, 3, 4] k=2 → 3. On [4, 1, 9] k=2 → 2. On [7] k=2 → 0.
a is nonempty. k ≥ 1. When k > n the window count is 0.