0075MediumStringDynamic ProgrammingBacktrackingBracket Sequences

Ridge Walk Plans

Tracked in this browser only

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 trail crew is drafting a loop hike along a ridge. The hike is a sequence of steps written as a string, where "U" gains one metre of height and "D" loses one metre.

A plan is valid when all three hold:

it uses exactly climbs up steps and exactly climbs down steps, so it is 2 * climbs steps long;
the hike ends back at the trailhead height;
after every single step the hiker is at or above the trailhead height, because the ridge drops away below that line.

Return every valid plan. The plans may be returned in any order.

Examples

Example 1

Input: climbs = 2
Output: ["UUDD", "UDUD"]

Either both climbs are taken before either descent, or the hiker rises and drops twice in a row. Starting with a descent is ruled out by the third condition.

Example 2

Input: climbs = 4
Output: ["UUUUDDDD", "UUUDUDDD", "UUUDDUDD", "UUUDDDUD", "UUDUUDDD", "UUDUDUDD", "UUDUDDUD", "UUDDUUDD", "UUDDUDUD", "UDUUUDDD", "UDUUDUDD", "UDUUDDUD", "UDUDUUDD", "UDUDUDUD"]

There are fourteen plans of eight steps that stay at or above the trailhead height and finish level with it.

Constraints

1 <= climbs <= 8

The values you return may be in any order.

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 ridge_walk_plans(climbs: int) -> list[str]:

Java

public List<String> ridgeWalkPlans(int climbs)

The intended approach

Time O(4^n / sqrt(n)) for n = climbs, proportional to the number of plans producedSpace O(n) for the recursion and the step buffer, excluding the output

Build a plan one step at a time while carrying the number of up and down steps already placed. An up step is allowed while fewer than `climbs` of them are placed, and a down step is allowed only while the down count trails the up count, which is exactly the condition that keeps the hiker from dipping under the trailhead. A finished plan is recorded when the step count reaches `2 * climbs`, and each choice is undone before the next one is tried.

Pythonentry point: ridge_walk_plans

def ridge_walk_plans(climbs):
    total = 2 * climbs
    plans = []
    steps = []

    def extend(ups, downs):
        if len(steps) == total:
            plans.append("".join(steps))
            return
        if ups < climbs:
            steps.append("U")
            extend(ups + 1, downs)
            steps.pop()
        if downs < ups:
            steps.append("D")
            extend(ups, downs + 1)
            steps.pop()

    extend(0, 0)
    return plans

The obvious version

Enumerate all 2^(2n) step strings by treating each bit of a counter as one step, then keep the ones whose running height never goes negative and ends at zero.

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_ridge_walk_plans(climbs):
    total = 2 * climbs
    found = []
    for code in range(1 << total):
        plan = []
        for i in range(total):
            plan.append("U" if (code >> i) & 1 else "D")
        height = 0
        ok = True
        for step in plan:
            height += 1 if step == "U" else -1
            if height < 0:
                ok = False
                break
        if ok and height == 0:
            found.append("".join(plan))
    return found

How this problem was checked

Reference solution executed against 6 test cases — 6 passed
2 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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in