0304EasyMathSimulationNumber Theory

Single Digit Label Check

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Trains the technique from

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.

Every parcel that leaves the depot carries a tracking number num, and the printer stamps a one-digit check mark next to it.

The check mark is produced like this: add up the digits of num. If that total is still two digits or longer, add up its digits as well, and keep folding the number that way until only one digit is left. That surviving digit is the check mark.

Given num, return its check mark.

Examples

Example 1

Input: num = 4917
Output: 3

4 + 9 + 1 + 7 = 21, and 2 + 1 = 3. Three is a single digit, so the check mark is 3.

Example 2

Input: num = 99
Output: 9

9 + 9 = 18, then 1 + 8 = 9.

Example 3

Input: num = 12345
Output: 6

The digits total 15, and 1 + 5 = 6.

Example 4

Input: num = 2147483647
Output: 1

The ten digits total 46, then 4 + 6 = 10, then 1 + 0 = 1. This is the largest tracking number the constraints allow.

Constraints

0 <= num <= 2^31 - 1

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 add_digits(num: int) -> int:

Java

public int addDigits(int num)

The intended approach

Time O(1)Space O(1)

Folding the digits of a number never changes its remainder modulo 9, because 10 leaves remainder 1. So the surviving digit is fixed by `num % 9`, with the single adjustment that a nonzero multiple of 9 folds down to 9 rather than 0. Writing that as `1 + (num - 1) % 9` handles every positive input, and 0 is returned unchanged.

Pythonentry point: add_digits

def add_digits(num):
    # O(1) time, O(1) space: arithmetic only, no loop over the digits.
    if num == 0:
        return 0
    return 1 + (num - 1) % 9

The obvious version

Literally simulate the folding: while the value has more than one digit, replace it with the sum of its decimal digits.

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_add_digits(num):
    cur = num
    while cur >= 10:
        total = 0
        while cur > 0:
            total += cur % 10
            cur //= 10
        cur = total
    return cur

How this problem was checked

Reference solution executed against 13 test cases — 13 passed
4 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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