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Relaying an Alert Down the Chain

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Trains the technique from

LeetCode 1376Time Needed to Inform All Employees

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 brigade has count members numbered 0 through count - 1. The chain of command is given as an array: supervisor[i] is the number of the member who supervises member i, except that supervisor[lead] == -1 because the lead answers to nobody. Following the links upwards from any member always arrives at the lead, so the chain of command forms a tree rooted there.

When a member learns of an alert, they spend relayTime[i] minutes passing it on, after which all of the members they directly supervise know it at the same moment. A member who supervises nobody has relayTime[i] == 0.

The lead learns of the alert at minute 0. Return the number of minutes that pass before every member of the brigade knows.

Examples

Example 1

Input: count = 7, lead = 0, supervisor = [-1, 0, 0, 1, 1, 2, 2], relayTime = [1, 2, 5, 0, 0, 0, 0]
Output: 6

Members 1 and 2 both hear the alert at minute 1. Members 3 and 4 hear it 2 minutes after that, at minute 3, and members 5 and 6 hear it 5 minutes after minute 1, at minute 6, by which point the whole brigade knows.

Example 2

Input: count = 4, lead = 0, supervisor = [-1, 0, 1, 2], relayTime = [3, 2, 1, 0]
Output: 6

The chain runs 0 to 1 to 2 to 3, so member 1 hears it at minute 3, member 2 at minute 5 and member 3 at minute 6.

Example 3

Input: count = 5, lead = 3, supervisor = [3, 3, 3, -1, 3], relayTime = [0, 0, 0, 0, 0]
Output: 0

The lead is member 3 and supervises the other four directly, spending no time on the relay, so everybody knows at minute 0.

Constraints

1 <= count <= 10^5
0 <= lead <= 10^5 - 1
supervisor.length == count
-1 <= supervisor[i] <= 10^5 - 1
relayTime.length == count
0 <= relayTime[i] <= 1000
lead is a member number, and supervisor[lead] is -1 while every other entry of supervisor is a member number.
Following the supervisor links from any member reaches the lead.

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 alert_minutes(count: int, lead: int, supervisor: list[int], relayTime: list[int]) -> int:

Java

public int alertMinutes(int count, int lead, int[] supervisor, int[] relayTime)

The intended approach

Time O(count): every member is pushed and popped once.Space O(count) for the report lists, the hearing times and the stack.

Turn the supervisor array into lists of direct reports, then walk down from the lead carrying the minute at which each member hears the alert: a direct report of `m` hears it `relayTime[m]` minutes after `m` did. The answer is the largest of those minutes. The walk is driven from an explicit stack, because the chain of command may be as deep as the brigade is large.

Pythonentry point: alert_minutes

def alert_minutes(count, lead, supervisor, relayTime):
    reports = [[] for _ in range(count)]
    for i in range(count):
        if i != lead:
            reports[supervisor[i]].append(i)
    heard_at = [0] * count
    # explicit stack: the chain of command can be as deep as the whole brigade,
    # so a recursive walk would run out of room
    stack = [lead]
    worst = 0
    while stack:
        member = stack.pop()
        when = heard_at[member]
        if when > worst:
            worst = when
        onward = when + relayTime[member]
        for junior in reports[member]:
            heard_at[junior] = onward
            stack.append(junior)
    return worst

The obvious version

For each member separately, climb the supervisor links up to the lead, adding the relay time of every supervisor passed, and keep the largest total found. Costs the depth of the chain per member.

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_alert_minutes(count, lead, supervisor, relayTime):
    worst = 0
    for i in range(count):
        total = 0
        cur = i
        while cur != lead:
            up = supervisor[cur]
            total += relayTime[up]
            cur = up
        if total > worst:
            worst = total
    return worst

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

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