0708MediumMathDynamic ProgrammingBacktracking

Clean Crate Stamps

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Trains the technique from

LeetCode 357Count Numbers with Unique Digits

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A depot stamps its crates with the whole numbers from 0 up to but not including 10^n. A stamp is clean when no digit occurs twice in it, written the ordinary way with no leading zeros: 9, 40 and 907 are clean, while 11, 121 and 900 are not.

Return how many stamps in that range are clean. When n is 0 the range holds the single number 0.

Examples

Example 1

Input: n = 1
Output: 10

The stamps run from 0 through 9. Each is a single digit, so none of them repeats a digit and all ten are clean.

Example 2

Input: n = 3
Output: 739

The stamps run from 0 through 999, and 739 of them use no digit twice. For instance 5, 40 and 907 count, while 11, 121 and 900 do not.

Example 3

Input: n = 5
Output: 32491

The stamps run from 0 through 99999, and 32491 of them use no digit twice. For instance 13579 counts, while 13577 does not.

Constraints

0 <= n <= 8

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def count_clean_stamps(n: int) -> int:

Java

public int countCleanStamps(int n)

The intended approach

Time O(n): one multiplication and one addition per digit length.Space O(1).

Count the clean stamps one digit length at a time rather than testing individual numbers. A stamp of a given length picks a non-zero first digit and then, for each later position, any digit not already used, so the count for each length is the previous length's count multiplied by the number of digits still unused. Carrying that product forward and adding it once per length, starting from the single clean stamp `0`, gives the answer in a few multiplications.

Pythonentry point: count_clean_stamps

def count_clean_stamps(n):
    # O(n): one running product per digit length, no enumeration.
    total = 1          # the stamp 0 is clean on its own
    ways = 9           # clean stamps of the current digit length
    spare = 9          # digits still unused for the next position
    for _ in range(n):
        total += ways
        ways *= spare
        spare -= 1
    return total

The obvious version

Walk every number in the range, turn it into text and check whether its digits are all different. Costs one test per stamp, so it grows by a factor of ten with each increase in n.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_count_clean_stamps(n):
    total = 0
    for x in range(10 ** n):
        text = str(x)
        if len(set(text)) == len(text):
            total += 1
    return total

How this problem was checked

Reference solution executed against 9 test cases — 9 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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