Order After K Digit Passes

Problem

A radix sort on whole numbers that are zero or more works one decimal digit at a time, starting from the ones digit. Each pass stably regroups the list by the current digit, 0 first and 9 last, keeping the existing order inside each group. Given a list and a number k, return the list exactly as it looks after the first k passes.

Examples

Input:  nums = [53, 7, 318, 90, 41, 206, 17], k = 1
Output: [90, 41, 53, 206, 7, 17, 318]
Why:    grouped by ones digit: 0, 1, 3, 6, 7, 7, 8; 7 stays ahead of 17

Input:  nums = [53, 7, 318, 90, 41, 206, 17], k = 2
Output: [206, 7, 17, 318, 41, 53, 90]
Why:    the second pass regroups the first pass's output by the tens digit

Input:  nums = [30, 3, 300], k = 0
Output: [30, 3, 300]
Why:    edge case, no passes, so nothing moves

Hints

0 / 3

1. Hint 1reveal
2. Hint 2locked
3. Hint 3locked

def after_passes(nums, k):
    order, place = list(nums), 1
    for _ in range(k):
        buckets = [[] for _ in range(10)]
        for x in order:                          # list order in, so the pass is stable
            buckets[x // place % 10].append(x)
        order = [x for b in buckets for x in b]  # read back digit 0 to 9
        place *= 10
    return order

print(after_passes([53, 7, 318, 90, 41, 206, 17], 1))  # -> [90, 41, 53, 206, 7, 17, 318]
print(after_passes([53, 7, 318, 90, 41, 206, 17], 2))  # -> [206, 7, 17, 318, 41, 53, 90]
print(after_passes([30, 3, 300], 0))                   # -> [30, 3, 300]