Rolling Price Minimum and Maximum

Challenge

For every complete consecutive window of exactly width observations, return [minimum_price, maximum_price]. Output windows from oldest to newest. Unlike the time-window-average exercise, width counts observations, not elapsed time. Return [] when fewer than width prices exist.

Input and output contract

0 ≤ n ≤ 50,000; positive integer width ≤ 60,000; integer prices 0..1,000,000. Test inputs have at most 2,000 complete windows, so max(0,n-width+1)≤2,000; this bounds result size. Duplicates remain valid observations.

All tests follow this contract. You do not need to validate malformed input. Return the requested result from rolling_price_window; printed output is not your answer. Standard-library tools are allowed. Implement your solution in Python 3.12.

Target complexity: O(n) time; O(min(n, width)) auxiliary space plus output

Return values are checked by type. Use the exact function name. Print output is not your answer. Tests have time and memory limits.

Public examples

[[4, 2, 2, 5, 1], 3]

[[2, 4], [2, 5], [1, 5]]

[[8, 2], 3]

[]

[[3, 3], 1]

[[3, 3], [3, 3]]

The worked solution unlocks after completion, or with Premium access.