0041EasyMathDynamic ProgrammingMemoization

Rung Ladder Climbs

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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 maintenance robot waits on the ground under a ladder whose rungs are numbered 1 through n, counting upward from the bottom. Each pull of its winch raises the robot by exactly one rung or by exactly two rungs, and it never lowers itself. The job is finished the moment the robot stands on rung n, and it must not overshoot that rung.

Two climbs are counted separately when their lists of pull sizes differ at some position, so a one-rung pull followed by a two-rung pull is a different climb from a two-rung pull followed by a one-rung pull.

Return how many climbs finish on rung n.

Examples

Example 1

Input: n = 1
Output: 1

A single one-rung pull is the only way up, since a two-rung pull would carry the robot past the top.

Example 2

Input: n = 5
Output: 8

Writing pulls in order: 1+1+1+1+1, 1+1+1+2, 1+1+2+1, 1+2+1+1, 2+1+1+1, 1+2+2, 2+1+2 and 2+2+1.

Example 3

Input: n = 7
Output: 21

The climbs that end on rung 7 split into those whose last pull was one rung (13 of them, ending from rung 6) and those whose last pull was two rungs (8 of them, ending from rung 5).

Constraints

1 <= n <= 45

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 ladder_climbs(n: int) -> int:

Java

public int ladderClimbs(int n)

The intended approach

Time O(n)Space O(1)

A climb that ends on rung k had a last pull of one rung, arriving from rung k-1, or of two rungs, arriving from rung k-2, and those two groups share no climb. So the count for rung k is the sum of the counts for the two rungs below it, with one climb reaching rung 1 and one way of standing on the ground. Sweep upward holding only the last two counts.

Pythonentry point: ladder_climbs

def ladder_climbs(n):
    ground, first = 1, 1
    for _ in range(n - 1):
        ground, first = first, ground + first
    return first

The obvious version

Unmemoised recursion that walks the decision tree of pulls from the top down: the count for a remaining distance is the count for one rung less plus the count for two rungs less, with a remaining distance below zero contributing nothing. Exponential, but it never uses the closed recurrence as a loop.

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_ladder_climbs(n):
    if n < 0:
        return 0
    if n <= 1:
        return 1
    return bf_ladder_climbs(n - 1) + bf_ladder_climbs(n - 2)

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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