LeetCode #1376 — MEDIUM

Time Needed to Inform All Employees

Move from brute-force thinking to an efficient approach using tree strategy.

The Problem

Problem Statement

A company has n employees with a unique ID for each employee from 0 to n - 1. The head of the company is the one with headID.

Each employee has one direct manager given in the manager array where manager[i] is the direct manager of the i-th employee, manager[headID] = -1. Also, it is guaranteed that the subordination relationships have a tree structure.

The head of the company wants to inform all the company employees of an urgent piece of news. He will inform his direct subordinates, and they will inform their subordinates, and so on until all employees know about the urgent news.

The i-th employee needs informTime[i] minutes to inform all of his direct subordinates (i.e., After informTime[i] minutes, all his direct subordinates can start spreading the news).

Return the number of minutes needed to inform all the employees about the urgent news.

Example 1:

Input: n = 1, headID = 0, manager = [-1], informTime = [0]
Output: 0
Explanation: The head of the company is the only employee in the company.

Example 2:

Input: n = 6, headID = 2, manager = [2,2,-1,2,2,2], informTime = [0,0,1,0,0,0]
Output: 1
Explanation: The head of the company with id = 2 is the direct manager of all the employees in the company and needs 1 minute to inform them all.
The tree structure of the employees in the company is shown.

Constraints:

1 <= n <= 10⁵
0 <= headID < n
manager.length == n
0 <= manager[i] < n
manager[headID] == -1
informTime.length == n
0 <= informTime[i] <= 1000
informTime[i] == 0 if employee i has no subordinates.
It is guaranteed that all the employees can be informed.

Step 01

Brute Force Baseline

Problem summary: A company has n employees with a unique ID for each employee from 0 to n - 1. The head of the company is the one with headID. Each employee has one direct manager given in the manager array where manager[i] is the direct manager of the i-th employee, manager[headID] = -1. Also, it is guaranteed that the subordination relationships have a tree structure. The head of the company wants to inform all the company employees of an urgent piece of news. He will inform his direct subordinates, and they will inform their subordinates, and so on until all employees know about the urgent news. The i-th employee needs informTime[i] minutes to inform all of his direct subordinates (i.e., After informTime[i] minutes, all his direct subordinates can start spreading the news). Return the number of minutes needed to inform all the employees about the urgent news.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Tree

Example 1

1
0
[-1]
[0]

Example 2

6
2
[2,2,-1,2,2,2]
[0,0,1,0,0,0]

Core Insight

What unlocks the optimal approach

The company can be represented as a tree, headID is always the root.
Store for each node the time needed to be informed of the news.
Answer is the max time a leaf node needs to be informed.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1376: Time Needed to Inform All Employees
class Solution {
    private List<Integer>[] g;
    private int[] informTime;

    public int numOfMinutes(int n, int headID, int[] manager, int[] informTime) {
        g = new List[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        this.informTime = informTime;
        for (int i = 0; i < n; ++i) {
            if (manager[i] >= 0) {
                g[manager[i]].add(i);
            }
        }
        return dfs(headID);
    }

    private int dfs(int i) {
        int ans = 0;
        for (int j : g[i]) {
            ans = Math.max(ans, dfs(j) + informTime[i]);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1376: Time Needed to Inform All Employees
func numOfMinutes(n int, headID int, manager []int, informTime []int) int {
	g := make([][]int, n)
	for i, x := range manager {
		if x != -1 {
			g[x] = append(g[x], i)
		}
	}
	var dfs func(int) int
	dfs = func(i int) (ans int) {
		for _, j := range g[i] {
			ans = max(ans, dfs(j)+informTime[i])
		}
		return
	}
	return dfs(headID)
}

# Accepted solution for LeetCode #1376: Time Needed to Inform All Employees
class Solution:
    def numOfMinutes(
        self, n: int, headID: int, manager: List[int], informTime: List[int]
    ) -> int:
        def dfs(i: int) -> int:
            ans = 0
            for j in g[i]:
                ans = max(ans, dfs(j) + informTime[i])
            return ans

        g = defaultdict(list)
        for i, x in enumerate(manager):
            g[x].append(i)
        return dfs(headID)

// Accepted solution for LeetCode #1376: Time Needed to Inform All Employees
struct Solution;

impl Solution {
    fn num_of_minutes(n: i32, _: i32, mut manager: Vec<i32>, mut inform_time: Vec<i32>) -> i32 {
        let n = n as usize;
        let mut res = 0;
        for i in 0..n {
            res = res.max(Self::dfs(i, &mut manager, &mut inform_time));
        }
        res
    }

    fn dfs(i: usize, manager: &mut Vec<i32>, inform_time: &mut Vec<i32>) -> i32 {
        if manager[i] != -1 {
            inform_time[i] += Self::dfs(manager[i] as usize, manager, inform_time);
            manager[i] = -1;
        }
        inform_time[i]
    }
}

#[test]
fn test() {
    let n = 1;
    let head_id = 0;
    let manager = vec![-1];
    let inform_time = vec![0];
    let res = 0;
    assert_eq!(
        Solution::num_of_minutes(n, head_id, manager, inform_time),
        res
    );
    let n = 6;
    let head_id = 2;
    let manager = vec![2, 2, -1, 2, 2, 2];
    let inform_time = vec![0, 0, 1, 0, 0, 0];
    let res = 1;
    assert_eq!(
        Solution::num_of_minutes(n, head_id, manager, inform_time),
        res
    );
    let n = 7;
    let head_id = 6;
    let manager = vec![1, 2, 3, 4, 5, 6, -1];
    let inform_time = vec![0, 6, 5, 4, 3, 2, 1];
    let res = 21;
    assert_eq!(
        Solution::num_of_minutes(n, head_id, manager, inform_time),
        res
    );
    let n = 15;
    let head_id = 0;
    let manager = vec![-1, 0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6];
    let inform_time = vec![1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0];
    let res = 3;
    assert_eq!(
        Solution::num_of_minutes(n, head_id, manager, inform_time),
        res
    );
    let n = 4;
    let head_id = 2;
    let manager = vec![3, 3, -1, 2];
    let inform_time = vec![0, 0, 162, 914];
    let res = 1076;
    assert_eq!(
        Solution::num_of_minutes(n, head_id, manager, inform_time),
        res
    );
}

// Accepted solution for LeetCode #1376: Time Needed to Inform All Employees
function numOfMinutes(n: number, headID: number, manager: number[], informTime: number[]): number {
    const g: number[][] = new Array(n).fill(0).map(() => []);
    for (let i = 0; i < n; ++i) {
        if (manager[i] !== -1) {
            g[manager[i]].push(i);
        }
    }
    const dfs = (i: number): number => {
        let ans = 0;
        for (const j of g[i]) {
            ans = Math.max(ans, dfs(j) + informTime[i]);
        }
        return ans;
    };
    return dfs(headID);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

LEVEL ORDER

O(n) time

O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL

O(n) time

O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.