0231EasyArrayDynamic Programming

Ladder Rung Repair Fees

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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 crew has to get up and over a service ladder whose rungs each need patching before they will take weight. The array fees holds the patching charge for every rung, counted from the bottom, so fees[i] is billed the moment a climber rests on rung i.

A climber opens the ascent by resting on rung 0 or on rung 1, whichever they prefer, and is billed for that rung. From whatever rung they are on, the next move lifts them one rung or two rungs higher. The ascent is over as soon as a move carries the climber past the highest rung, and clearing the ladder is not billed.

Return the smallest total charge for an ascent that gets past the top.

Examples

Example 1

Input: fees = [9, 1, 1]
Output: 1

The climber opens on rung 1 and is billed 1, then a two-rung move carries them past rung 2, which is the top of this ladder. Nothing more is billed.

Example 2

Input: fees = [1, 2, 3, 100, 1]
Output: 5

Resting on rungs 0, 2 and 4 is billed 1 + 3 + 1 = 5, and one more move from rung 4 clears the ladder.

Example 3

Input: fees = [4, 3, 2, 6]
Output: 5

Resting on rung 1 and then rung 2 is billed 3 + 2 = 5, and a two-rung move from rung 2 carries the climber past rung 3.

Constraints

2 <= fees.length <= 1000
0 <= fees[i] <= 999

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 cheapest_rung_climb(fees: list[int]) -> int:

Java

public int cheapestRungClimb(int[] fees)

The intended approach

Time O(n)Space O(1)

Work with the smallest charge that gets a climber to each point on the way up, where reaching point i means being poised at rung i without having rested on it, and point n means clear of the ladder. Points 0 and 1 cost nothing because the ascent may open on either of the bottom two rungs, and any later point is reached by resting on the rung one below it or the rung two below it and paying that rung's fee. Only the two previous figures are ever consulted, so two scalars roll up the ladder and the figure for point n is returned.

Pythonentry point: cheapest_rung_climb

def cheapest_rung_climb(fees):
    below = 0
    above = 0
    for i in range(2, len(fees) + 1):
        cheapest = min(above + fees[i - 1], below + fees[i - 2])
        below = above
        above = cheapest
    return above

The obvious version

Enumerate every possible ascent with an explicit stack, seeded with both opening rungs, adding each rung's fee as it is rested on and keeping the cheapest run that can clear the top. Exponential in the number of rungs.

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_cheapest_rung_climb(fees):
    n = len(fees)
    best = None
    stack = [(0, 0), (1, 0)]
    while stack:
        rung, paid = stack.pop()
        paid += fees[rung]
        if rung + 1 >= n or rung + 2 >= n:
            if best is None or paid < best:
                best = paid
        if rung + 1 < n:
            stack.append((rung + 1, paid))
        if rung + 2 < n:
            stack.append((rung + 2, paid))
    return best

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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