0995EasyLinked ListMath

Reading a Punch Chain as a Number

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Trains the technique from

LeetCode 1290Convert Binary Number in a Linked List to 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.

A tally machine punches cards and clips them one behind another. Every card carries a single hole state, either 0 or 1, plus a clip fastened to the card behind it, and the card at the end has an empty clip. Nothing in the machine records how many cards a chain holds.

Because the harness passes plain JSON, the chain reaches you as punches, listing the hole states in clip order starting at the front card. There is always at least one card.

Read the chain as a number written in base two, with the front card carrying the largest place value and the card at the end carrying the ones. Return that number.

Examples

Example 1

Input: punches = [1, 1, 0]
Output: 6

The front card is worth four, the middle one two, and the card at the end nothing, so the chain reads six.

Example 2

Input: punches = [1, 0, 0, 1]
Output: 9

Only the front card and the card at the end are punched, worth eight and one, which adds to nine.

Example 3

Input: punches = [0, 0, 1]
Output: 1

The leading unpunched cards carry no weight, so only the ones place counts.

Constraints

1 <= punches.length <= 30
0 <= punches[i] <= 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 get_decimal_value(punches: list[int]) -> int:

Java

public int getDecimalValue(int[] punches)

The intended approach

Time O(n) over the chain's length.Space O(1).

Walk the chain from the front, carrying the number read so far. At each card, double what has been read and add that card's hole state. Doubling is right because moving one card further along pushes every place already read up by one, and the new card takes the ones place. That needs no knowledge of the chain's length ahead of time, which is the point when only the clips can be followed: there is no way to know a card's place value until the end is reached, but doubling as you go never asks. One pass, one running number, nothing else held.

Pythonentry point: get_decimal_value

def get_decimal_value(punches):
    total = 0
    for state in punches:
        total = total * 2 + state
    return total

The obvious version

Walk the chain once to count the cards, then walk it again adding the place value each card is worth, counting the places down from the front. Needs the length up front and works out a power of two per card, rather than doubling as it goes.

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_get_decimal_value(punches):
    length = 0
    for _state in punches:
        length += 1
    total = 0
    place = length - 1
    for state in punches:
        if state:
            total += 2 ** place
        place -= 1
    return total

How this problem was checked

Reference solution executed against 14 test cases — 14 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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