0413MediumBacktrackingTreeDepth-First SearchBinary Tree

Ridge Descents Matching a Drop

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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 survey of a mountain ridge records a network of marked waypoints. The topmost waypoint is the saddle, and every waypoint sends at most two marked forks further down the ridge, one bearing left and one bearing right. Each waypoint carries a signed height change in metres, positive where the ground rises onto it and negative where it falls onto it.

The survey arrives as waypoints, which lists the network one depth band at a time, left to right. Slot 0 holds the saddle. Each listed waypoint takes the next two unclaimed slots for its downhill forks, the left fork first and the right fork second. A slot holding null means no waypoint hangs there and it claims no slots of its own. Trailing null slots may be dropped from the end of the listing, and an empty listing means the survey found nothing.

A descent starts at the saddle, follows marked forks downhill, and stops at a waypoint with no fork leaving it. Its drop is the total of the height changes of every waypoint it touches, the saddle included.

Return every descent whose drop equals targetSum, each one given as the list of height changes in the order they are walked, from the saddle downward. The descents may come back in any order. Return an empty list when no descent matches.

Examples

Example 1

Input: waypoints = [6, 4, 9, 2, -1, null, 5], targetSum = 12
Output: [[6, 4, 2]]

The saddle records 6. Its left fork reaches 4, whose left fork holds 2, and that waypoint has nothing below it: 6 + 4 + 2 = 12. The survey's other two descents finish at -1 and at 5, dropping 9 and 20.

Example 2

Input: waypoints = [3, 5, null, -2, null, 4, null], targetSum = 10
Output: [[3, 5, -2, 4]]

The ridge runs straight down: 3, then 5, then -2, then 4. Only the last of those has no fork leaving it, and the four height changes total 10.

Example 3

Input: waypoints = [7], targetSum = 7
Output: [[7]]

The saddle has no fork, so the single descent touches only it and drops 7, which matches.

Example 4

Input: waypoints = [], targetSum = 0
Output: []

The survey lists no waypoints at all, so there is no descent to report.

Example 5

Input: waypoints = [4, 1, 2, null, null, 6, -1], targetSum = 5
Output: [[4, 1], [4, 2, -1]]

The saddle's left fork holds 1 with nothing below it, and 4 + 1 = 5. Its right fork holds 2 and still has two forks of its own: the one holding -1 brings that descent back to 5, while the one holding 6 finishes at 12.

Constraints

The number of listed waypoints is in the range [0, 5000].
-1000 <= waypoints[i] <= 1000 for every non-null slot
-1000 <= targetSum <= 1000
waypoints[0] is not null unless the listing is empty
waypoints is a valid depth-band listing: every non-null slot after slot 0 was claimed as a fork by an earlier waypoint

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 path_sum(waypoints: list[int | None], targetSum: int) -> list[list[int]]:

Java

public List<List<Integer>> pathSum(Integer[] waypoints, int targetSum)

The intended approach

Time O(m + s) where m is the listing length and s is the total number of height changes across the returned descents: each slot is entered and left once, and only a matching chain is copied.Space O(m + s): the two fork tables and the stack are proportional to the listing, the chain is bounded by the ridge depth, and the returned descents hold s numbers.

One sweep of the listing hands each waypoint the slot numbers of its two forks, honouring the rule that a null slot claims nothing. A second sweep walks the network from an explicit stack that carries an enter marker and a leave marker per waypoint: entering appends the height change to the current chain and to a running total, leaving strips both back off. At a waypoint with no fork the running total is compared with the target and a fresh copy of the chain is recorded on a hit. Driving the walk from a stack rather than nested calls keeps a 2500-deep ridge safe.

Pythonentry point: path_sum

def path_sum(waypoints, targetSum):
    # Resolve the listing into child slots, then walk it from an explicit stack
    # so a 2500-deep ridge cannot blow the interpreter's frame limit. The
    # running total and the current chain are pushed and popped as the walk
    # moves, and a chain is copied out only when a dead end matches.
    size = len(waypoints)
    if size == 0 or waypoints[0] is None:
        return []
    left = [-1] * size
    right = [-1] * size
    free = 1
    for idx in range(size):
        if waypoints[idx] is None:
            continue
        if free < size:
            if waypoints[free] is not None:
                left[idx] = free
            free += 1
        if free < size:
            if waypoints[free] is not None:
                right[idx] = free
            free += 1

    found = []
    chain = []
    running = 0
    stack = [(0, False)]
    while stack:
        idx, leaving = stack.pop()
        if leaving:
            running -= chain.pop()
            continue
        value = waypoints[idx]
        chain.append(value)
        running += value
        stack.append((idx, True))
        if left[idx] < 0 and right[idx] < 0:
            if running == targetSum:
                found.append(chain[:])
        else:
            if right[idx] >= 0:
                stack.append((right[idx], False))
            if left[idx] >= 0:
                stack.append((left[idx], False))
    return found

The obvious version

Give every waypoint the slot number of the waypoint above it, then take each fork-free waypoint in turn, climb slot by slot up to the saddle, reverse what was collected, and keep the chain when its plain sum equals the target.

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_path_sum(waypoints, targetSum):
    size = len(waypoints)
    if size == 0 or waypoints[0] is None:
        return []
    owner = [-1] * size
    kids = [0] * size
    free = 1
    for idx in range(size):
        if waypoints[idx] is None:
            continue
        for _ in range(2):
            if free < size:
                if waypoints[free] is not None:
                    owner[free] = idx
                    kids[idx] += 1
                free += 1
    answer = []
    for idx in range(size):
        if waypoints[idx] is None or kids[idx] != 0:
            continue
        chain = []
        here = idx
        while here != -1:
            chain.append(waypoints[here])
            here = owner[here]
        chain.reverse()
        if sum(chain) == targetSum:
            answer.append(chain)
    return answer

How this problem was checked

Reference solution executed against 13 test cases — 13 passed
5 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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