0298MediumHash TableTreeDepth-First SearchBreadth-First SearchBinary Tree

Firmware Cascade Minutes

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Trains the technique from

LeetCode 2385Amount of Time for Binary Tree to Be Infected

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 rack of relays is wired as a binary tree. The wiring arrives as layout, the level-order walk of that tree: layout[0] is the root relay, and the walk then lists the left and right child slot of every relay it has already listed, in the same order, writing null for a slot with no relay in it. A null slot has no child slots of its own, and trailing null entries are left off the end. Every relay carries a different label.

A firmware update is loaded onto the relay labelled seed. Each wire joins a relay to a child and carries the update in either direction. During one minute, every relay that already holds the update pushes it across all of its wires, so each neighbouring relay without the update picks it up.

Return the number of minutes until every relay in the rack holds the update. A rack of one relay is already done, so the answer is 0.

Examples

Example 1

Input: layout = [4,2,9,1,3,null,7], seed = 2
Output: 3

Minute 1 puts the update on relays 1, 3 and 4. Minute 2 puts it on relay 9, and minute 3 puts it on relay 7, which is the last relay in the rack.

Example 2

Input: layout = [8,3,10,1,6,null,14,null,null,4,7,13], seed = 6
Output: 5

The update spreads to relays 3, 4 and 7 in minute 1, to 1 and 8 in minute 2, to 10 in minute 3, to 14 in minute 4 and to 13 in minute 5.

Constraints

The number of relays in the rack is in the range [1, 10^5].
1 <= relay label <= 10^5
Every relay carries a different label.
layout is the level-order walk of the tree, with null for an empty child slot and trailing nulls omitted.
seed is the label of a relay that is present in the rack.

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 cascade_minutes(layout: list[int | None], seed: int) -> int:

Java

public int cascadeMinutes(Integer[] layout, int seed)

The intended approach

Time O(n)Space O(n)

Replay the level-order walk with a queue to recover which relays are wired together, recording every wire in both directions so a parent is reachable from a child. Then sweep outward from the seed with a second queue that stamps a hop count on each newly reached relay, and return the largest hop count seen.

Pythonentry point: cascade_minutes

from collections import deque


def cascade_minutes(layout, seed):
    # Rebuild the links from the level-order walk, keeping each link in both
    # directions, then sweep outward from the seed with a queue.  The furthest hop
    # count reached is the number of minutes.
    # O(n) time, O(n) space; the sweep is queue driven, so depth is never a concern.
    links = {}
    for value in layout:
        if value is not None:
            links[value] = []
    total = len(layout)
    waiting = deque([layout[0]])
    idx = 1
    while waiting and idx < total:
        parent = waiting.popleft()
        for _ in range(2):
            if idx >= total:
                break
            child = layout[idx]
            idx += 1
            if child is not None:
                links[parent].append(child)
                links[child].append(parent)
                waiting.append(child)

    seen = {seed}
    frontier = deque([(seed, 0)])
    furthest = 0
    while frontier:
        node, hops = frontier.popleft()
        if hops > furthest:
            furthest = hops
        for nxt in links[node]:
            if nxt not in seen:
                seen.add(nxt)
                frontier.append((nxt, hops + 1))
    return furthest

The obvious version

Recover only the parent of each relay, then for every relay walk its ancestor chain until it meets the seed's ancestor chain and add the two chain offsets to get the distance, keeping the largest.

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_cascade_minutes(layout, seed):
    root = layout[0]
    parent = {root: None}
    order = [root]
    head = 0
    idx = 1
    total = len(layout)
    while head < len(order) and idx < total:
        p = order[head]
        head += 1
        for _ in range(2):
            if idx >= total:
                break
            child = layout[idx]
            idx += 1
            if child is not None:
                parent[child] = p
                order.append(child)

    def chain(x):
        path = []
        while x is not None:
            path.append(x)
            x = parent[x]
        return path

    seed_pos = {v: i for i, v in enumerate(chain(seed))}
    best = 0
    for v in parent:
        for i, a in enumerate(chain(v)):
            if a in seed_pos:
                d = i + seed_pos[a]
                break
        if d > best:
            best = d
    return best

How this problem was checked

Reference solution executed against 9 test cases — 9 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

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