0719EasyMath

Crate Serial Checksum

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Trains the technique from

LeetCode 1281Subtract the Product and Sum of Digits of an Integer

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 crate leaving the yard is stamped with a serial number serial, a positive whole number written in base ten with no leading zeros.

The yard's clerk works out a checksum for the crate by multiplying all the decimal digits of serial together, adding all those same digits up, and then subtracting the second tally from the first.

Return the checksum. It may be negative.

Examples

Example 1

Input: serial = 705
Output: -12

The digits are 7, 0 and 5. Their product is 0 and their sum is 12, so the checksum is 0 - 12.

Example 2

Input: serial = 48
Output: 20

The digits are 4 and 8. Their product is 32 and their sum is 12, so the checksum is 32 - 12.

Example 3

Input: serial = 9
Output: 0

A one-digit serial has a product of 9 and a sum of 9, so the checksum is 9 - 9.

Constraints

1 <= serial <= 10^5

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 digit_checksum(serial: int) -> int:

Java

public int digitChecksum(int serial)

The intended approach

Time O(d) for a serial with d decimal digits.Space O(1)

Take the serial apart one decimal digit at a time: dividing by ten gives the remaining digits and the remainder gives the digit just removed. Multiply each digit into a product that starts at one and add it into a sum that starts at zero, then return the product minus the sum.

Pythonentry point: digit_checksum

def digit_checksum(serial):
    # Peel one decimal digit at a time with divmod, keeping a running product and
    # a running sum, then report their difference.
    # Time O(d) for a serial of d digits.  Space O(1).
    product = 1
    total = 0
    left = serial
    while left > 0:
        left, digit = divmod(left, 10)
        product *= digit
        total += digit
    return product - total

The obvious version

Write the serial out as decimal text, turn each character into its digit, then take one pass to multiply the list and a second pass to add it up before subtracting.

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_digit_checksum(serial):
    digits = [int(ch) for ch in str(serial)]
    product = 1
    for d in digits:
        product = product * d
    total = 0
    for d in digits:
        total = total + d
    return product - total

How this problem was checked

Reference solution executed against 10 test cases — 10 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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