0110EasyTreeDepth-First SearchBreadth-First SearchBinary Tree

Longest Run Through the Delta

Tracked in this browser only

Trains the technique from

LeetCode 104Maximum Depth of Binary Tree

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 charts a river delta as junctions that split the flow downstream. Each junction sends water onward through at most two channels, one toward the west bank and one toward the east bank, and each junction records a signed bed grade in centimetres per kilometre.

The chart arrives as junctions, a listing that walks the delta one tier at a time starting at the source, with the west entry of a tier written before its east entry. Slot 0 holds the source junction. Every junction that appears in the listing claims the next two free slots, its west channel first and its east channel second. A slot holding null means no channel leaves in that direction, and such a slot claims no slots of its own. Trailing null slots may be left off, and an empty listing means nothing was charted.

A run begins at the source junction, follows channels downstream only, and halts at a junction that sends water nowhere. Report how many junctions sit on the longest run. When nothing was charted, report 0.

Examples

Example 1

Input: junctions = [12, -5, 40, null, 7]
Output: 3

The source splits west into -5 and east into 40. The junction -5 sends water east into 7, so the longest run visits three junctions while the run through 40 visits only two.

Example 2

Input: junctions = []
Output: 0

Nothing was charted, so no run exists and the answer is zero.

Example 3

Input: junctions = [-100]
Output: 1

The source sends water nowhere, so the only run halts immediately after visiting one junction.

Constraints

The number of junctions in the chart is in the range [0, 10^4].
-100 <= grade of a junction <= 100
An empty listing means no junction was charted.

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 delta_run_length(junctions: list[int | None]) -> int:

Java

public int deltaRunLength(Integer[] junctions)

The intended approach

Time O(n)Space O(n)

Rebuild the west and east channel links from the tier-by-tier listing, then sweep the delta one tier at a time starting from the source. Each sweep replaces the current tier with every junction it feeds, so the number of sweeps performed before the tier comes back empty is exactly the count of junctions on the longest run. An empty chart never enters the loop and answers zero.

Pythonentry point: delta_run_length

def delta_run_length(junctions):
    n = len(junctions)
    if n == 0 or junctions[0] is None:
        return 0
    west = [-1] * n
    east = [-1] * n
    listed = [0]
    slot = 1
    head = 0
    while head < len(listed):
        cur = listed[head]
        head += 1
        if slot < n:
            if junctions[slot] is not None:
                west[cur] = slot
                listed.append(slot)
            slot += 1
        if slot < n:
            if junctions[slot] is not None:
                east[cur] = slot
                listed.append(slot)
            slot += 1
    tier = [0]
    reached = 0
    while tier:
        reached += 1
        below = []
        for idx in tier:
            if west[idx] >= 0:
                below.append(west[idx])
            if east[idx] >= 0:
                below.append(east[idx])
        tier = below
    return reached

The obvious version

Record which junction feeds each junction, then for every junction climb back upstream to the source one step at a time and count the junctions passed. The largest count wins. O(n^2).

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_delta_run_length(junctions):
    n = len(junctions)
    if n == 0 or junctions[0] is None:
        return 0
    upstream = {0: -1}
    listed = [0]
    slot = 1
    head = 0
    while head < len(listed):
        cur = listed[head]
        head += 1
        for _ in range(2):
            if slot < n:
                if junctions[slot] is not None:
                    upstream[slot] = cur
                    listed.append(slot)
                slot += 1
    longest = 0
    for idx in listed:
        steps = 1
        at = idx
        while upstream[at] != -1:
            at = upstream[at]
            steps += 1
        if steps > longest:
            longest = steps
    return longest

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

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