0118MediumLinked ListMathRecursion

Odometer Tape Sum

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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.

A workshop logs mileage counters onto punched paper tapes. A tape is a one-way chain of cells: every cell holds a single digit and links to the cell punched after it, and the final cell links to nothing. The punch head starts at the smallest place value, so the front cell of a tape carries the ones digit, the cell after it carries the tens digit, and so on.

Because the harness passes plain JSON, each tape arrives as an array of its digits in punch order, ones digit at index 0. You get two of them, left and right. Neither tape is empty, and neither carries useless high-order padding: the last cell of a tape is nonzero unless the whole tape is a single 0.

Return the tape that records the sum of the two counter values, laid out the same way: an array of digits with the ones digit first. Produce it cell by cell while stepping through both chains together and pushing any overflow into the next place. Do not fold either chain into one big number before adding.

Examples

Example 1

Input: left = [6, 8, 4], right = [7, 3]
Output: [3, 2, 5]

The tapes record 486 and 37. Adding the ones cells gives 13, so the answer's front cell is 3 and 1 rides into the tens place, where 8 + 3 + 1 = 12 leaves 2 and another rider, and 4 + 1 = 5 finishes the hundreds place.

Example 2

Input: left = [5, 9, 9], right = [8, 1]
Output: [3, 1, 0, 1]

995 plus 18 is 1013. The overflow survives past the end of both tapes, so the answer needs one more cell than the longer input.

Example 3

Input: left = [0], right = [4, 7]
Output: [4, 7]

A counter sitting at zero is punched as a single 0 cell, and adding it leaves 74 untouched.

Constraints

1 <= left.length <= 100
1 <= right.length <= 100
0 <= left[i] <= 9
0 <= right[i] <= 9
Neither tape has high-order padding: its last digit is nonzero unless the tape is exactly [0]

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_odometer_tapes(left: list[int], right: list[int]) -> list[int]:

Java

public int[] sumOdometerTapes(int[] left, int[] right)

The intended approach

Time O(n + m) where n and m are the tape lengthsSpace O(max(n, m)) for the chains built from the input and the answer chain, plus O(1) cursors beyond them

Materialise both chains of cells, then advance a cursor along each one in step while keeping a single overflow digit. At each place value add whichever cursors are still alive plus the overflow, append a fresh cell holding the remainder mod 10, and keep the tenth part as the next overflow. The walk continues while either cursor survives or an overflow is still pending, which is what grows the answer by one cell when the top place spills.

Pythonentry point: sum_odometer_tapes

def sum_odometer_tapes(left, right):
    class Cell:
        __slots__ = ("digit", "nxt")

        def __init__(self, digit):
            self.digit = digit
            self.nxt = None

    def build(tape):
        head = None
        tail = None
        for d in tape:
            cell = Cell(d)
            if head is None:
                head = cell
            else:
                tail.nxt = cell
            tail = cell
        return head

    a = build(left)
    b = build(right)
    out_head = None
    out_tail = None
    overflow = 0
    while a is not None or b is not None or overflow:
        total = overflow
        if a is not None:
            total += a.digit
            a = a.nxt
        if b is not None:
            total += b.digit
            b = b.nxt
        overflow = total // 10
        cell = Cell(total % 10)
        if out_head is None:
            out_head = cell
        else:
            out_tail.nxt = cell
        out_tail = cell
    answer = []
    node = out_head
    while node is not None:
        answer.append(node.digit)
        node = node.nxt
    return answer

The obvious version

Ignore the chain entirely: weigh each digit by its power of ten to recover both counter values, add them as ordinary integers, then peel the total apart with repeated division to rebuild the tape.

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_tape_sum(left, right):
    def value(tape):
        total = 0
        for i, d in enumerate(tape):
            total += d * (10 ** i)
        return total

    s = value(left) + value(right)
    if s == 0:
        return [0]
    digits = []
    while s > 0:
        digits.append(s % 10)
        s //= 10
    return digits

How this problem was checked

Reference solution executed against 8 test cases — 8 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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