0342EasyMath

Trimmed Meter Reading

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Trains the technique from

LeetCode 3754Concatenate Non-Zero Digits and Multiply by Sum I

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 water meter prints a whole-number reading n. The billing system turns that reading into a single audit figure from two ingredients.

The trimmed reading: walk the decimal digits of n from left to right, throw away every digit equal to 0, and read whatever digits are left as a decimal number, keeping their original order. When nothing is left, the trimmed reading is 0.
The digit total: the sum of all decimal digits of n.

Return the trimmed reading multiplied by the digit total. The product can be larger than a signed 32-bit integer holds.

Examples

Example 1

Input: n = 58306
Output: 128392

Dropping the single `0` leaves the digits `5`, `8`, `3`, `6`, so the trimmed reading is 58306 without its zero, namely 5836. The digit total is 5 + 8 + 3 + 0 + 6 = 22, and 5836 * 22 = 128392.

Example 2

Input: n = 4907
Output: 9940

The trimmed reading is 497 and the digit total is 4 + 9 + 0 + 7 = 20, so the answer is 497 * 20 = 9940.

Example 3

Input: n = 91
Output: 910

No digit is thrown away, so the trimmed reading equals the reading itself. The digit total is 9 + 1 = 10, and 91 * 10 = 910.

Constraints

0 <= n <= 10^9

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 sum_and_multiply(n: int) -> int:

Java

public long sumAndMultiply(int n)

The intended approach

Time O(d), where d is the number of decimal digits of nSpace O(1)

Peel digits off the units end with repeated division by ten. Every digit is added to the running total, and a non-zero digit is written into the trimmed reading at the next unused power of ten, which restores the original left-to-right order without any string work. A reading of `0` never enters the loop, so both ingredients stay zero.

Pythonentry point: sum_and_multiply

def sum_and_multiply(n):
    # One walk over the decimal digits, from the units end upward: accumulate the
    # digit total, and rebuild the trimmed reading by placing each surviving digit
    # at the next free power of ten.
    # Time O(d) for the d digits of n.  Space O(1).
    total = 0
    trimmed = 0
    place = 1
    rest = n
    while rest > 0:
        digit = rest % 10
        rest //= 10
        total += digit
        if digit != 0:
            trimmed += digit * place
            place *= 10
    return trimmed * total

The obvious version

Render the reading as text, filter the zero characters out of it, convert the surviving text back to an integer, and add up the characters of the original text for the total.

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_trimmed_product(n):
    text = str(n)
    kept = "".join(ch for ch in text if ch != "0")
    trimmed = int(kept) if kept else 0
    total = 0
    for ch in text:
        total = total + int(ch)
    return trimmed * 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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