Sliding Max - Algorithm

For each index $i$:

$1.$ Remove indices from front that are outside window ($< i - k + 1$).

$2.$ Remove indices from back while $nums[i] >= nums[deque.back()]$. These can never be the answer.

$3.$ Push $i$ to back.

$4.$ If $i >= k - 1$ (window is full), the front is the maximum. Each index is pushed once and popped at most once. Total: $O(n)$. Time: $O(n)$. Space: $O(k)$.