LeetCode #743 — MEDIUM

Network Delay Time

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

The Problem

Problem Statement

You are given a network of n nodes, labeled from 1 to n. You are also given times, a list of travel times as directed edges times[i] = (u_i, v_i, w_i), where u_i is the source node, v_i is the target node, and w_i is the time it takes for a signal to travel from source to target.

We will send a signal from a given node k. Return the minimum time it takes for all the n nodes to receive the signal. If it is impossible for all the n nodes to receive the signal, return -1.

Example 1:

Input: times = [[2,1,1],[2,3,1],[3,4,1]], n = 4, k = 2
Output: 2

Example 2:

Input: times = [[1,2,1]], n = 2, k = 1
Output: 1

Example 3:

Input: times = [[1,2,1]], n = 2, k = 2
Output: -1

Constraints:

1 <= k <= n <= 100
1 <= times.length <= 6000
times[i].length == 3
1 <= u_i, v_i <= n
u_i != v_i
0 <= w_i <= 100
All the pairs (u_i, v_i) are unique. (i.e., no multiple edges.)

Step 01

Brute Force Baseline

Problem summary: You are given a network of n nodes, labeled from 1 to n. You are also given times, a list of travel times as directed edges times[i] = (ui, vi, wi), where ui is the source node, vi is the target node, and wi is the time it takes for a signal to travel from source to target. We will send a signal from a given node k. Return the minimum time it takes for all the n nodes to receive the signal. If it is impossible for all the n nodes to receive the signal, return -1.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

[[2,1,1],[2,3,1],[3,4,1]]
4
2

Example 2

[[1,2,1]]
2
1

Example 3

[[1,2,1]]
2
2

Core Insight

What unlocks the optimal approach

We visit each node at some time, and if that time is better than the fastest time we've reached this node, we travel along outgoing edges in sorted order. Alternatively, we could use Dijkstra's algorithm.

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 #743: Network Delay Time
class Solution {
    public int networkDelayTime(int[][] times, int n, int k) {
        int[][] g = new int[n][n];
        int[] dist = new int[n];
        final int inf = 1 << 29;
        Arrays.fill(dist, inf);
        for (var e : g) {
            Arrays.fill(e, inf);
        }
        for (var e : times) {
            g[e[0] - 1][e[1] - 1] = e[2];
        }
        dist[k - 1] = 0;
        boolean[] vis = new boolean[n];
        for (int i = 0; i < n; ++i) {
            int t = -1;
            for (int j = 0; j < n; ++j) {
                if (!vis[j] && (t == -1 || dist[t] > dist[j])) {
                    t = j;
                }
            }
            vis[t] = true;
            for (int j = 0; j < n; ++j) {
                dist[j] = Math.min(dist[j], dist[t] + g[t][j]);
            }
        }
        int ans = 0;
        for (int x : dist) {
            ans = Math.max(ans, x);
        }
        return ans == inf ? -1 : ans;
    }
}

// Accepted solution for LeetCode #743: Network Delay Time
func networkDelayTime(times [][]int, n int, k int) int {
	const inf = 1 << 29
	g := make([][]int, n)
	for i := range g {
		g[i] = make([]int, n)
		for j := range g[i] {
			g[i][j] = inf
		}
	}
	for _, e := range times {
		g[e[0]-1][e[1]-1] = e[2]
	}

	dist := make([]int, n)
	for i := range dist {
		dist[i] = inf
	}
	dist[k-1] = 0

	vis := make([]bool, n)
	for i := 0; i < n; i++ {
		t := -1
		for j := 0; j < n; j++ {
			if !vis[j] && (t == -1 || dist[t] > dist[j]) {
				t = j
			}
		}
		vis[t] = true
		for j := 0; j < n; j++ {
			dist[j] = min(dist[j], dist[t]+g[t][j])
		}
	}

	if ans := slices.Max(dist); ans != inf {
		return ans
	}
	return -1
}

# Accepted solution for LeetCode #743: Network Delay Time
class Solution:
    def networkDelayTime(self, times: List[List[int]], n: int, k: int) -> int:
        g = [[inf] * n for _ in range(n)]
        for u, v, w in times:
            g[u - 1][v - 1] = w
        dist = [inf] * n
        dist[k - 1] = 0
        vis = [False] * n
        for _ in range(n):
            t = -1
            for j in range(n):
                if not vis[j] and (t == -1 or dist[t] > dist[j]):
                    t = j
            vis[t] = True
            for j in range(n):
                dist[j] = min(dist[j], dist[t] + g[t][j])
        ans = max(dist)
        return -1 if ans == inf else ans

// Accepted solution for LeetCode #743: Network Delay Time
struct Solution;
use std::collections::VecDeque;

impl Solution {
    fn network_delay_time(times: Vec<Vec<i32>>, n: i32, k: i32) -> i32 {
        let n = n as usize;
        let k = k as usize - 1;
        let mut graph: Vec<Vec<(usize, i32)>> = vec![vec![]; n];
        for time in times {
            let u = time[0] as usize - 1;
            let v = time[1] as usize - 1;
            let t = time[2];
            graph[u].push((v, t));
        }
        let mut visited = vec![std::i32::MAX; n];
        let mut queue = VecDeque::new();
        visited[k] = 0;
        queue.push_back(k);
        while let Some(u) = queue.pop_front() {
            for &(v, t) in &graph[u] {
                if t + visited[u] < visited[v] {
                    visited[v] = t + visited[u];
                    queue.push_back(v);
                }
            }
        }
        let max = visited.into_iter().max().unwrap();
        if max == std::i32::MAX {
            -1
        } else {
            max
        }
    }
}

#[test]
fn test() {
    let times = vec_vec_i32![[2, 1, 1], [2, 3, 1], [3, 4, 1]];
    let n = 4;
    let k = 2;
    let res = 2;
    assert_eq!(Solution::network_delay_time(times, n, k), res);
}

// Accepted solution for LeetCode #743: Network Delay Time
function networkDelayTime(times: number[][], n: number, k: number): number {
    const g: number[][] = Array.from({ length: n }, () => Array(n).fill(Infinity));
    for (const [u, v, w] of times) {
        g[u - 1][v - 1] = w;
    }
    const dist: number[] = Array(n).fill(Infinity);
    dist[k - 1] = 0;
    const vis: boolean[] = Array(n).fill(false);
    for (let i = 0; i < n; ++i) {
        let t = -1;
        for (let j = 0; j < n; ++j) {
            if (!vis[j] && (t === -1 || dist[j] < dist[t])) {
                t = j;
            }
        }
        vis[t] = true;
        for (let j = 0; j < n; ++j) {
            dist[j] = Math.min(dist[j], dist[t] + g[t][j]);
        }
    }
    const ans = Math.max(...dist);
    return ans === Infinity ? -1 : ans;
}

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^2 + m)

Space

O(n^2)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.