LeetCode #3112 — MEDIUM

Minimum Time to Visit Disappearing Nodes

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

The Problem

Problem Statement

There is an undirected graph of n nodes. You are given a 2D array edges, where edges[i] = [u_i, v_i, length_i] describes an edge between node u_i and node v_i with a traversal time of length_i units.

Additionally, you are given an array disappear, where disappear[i] denotes the time when the node i disappears from the graph and you won't be able to visit it.

Note that the graph might be disconnected and might contain multiple edges.

Return the array answer, with answer[i] denoting the minimum units of time required to reach node i from node 0. If node i is unreachable from node 0 then answer[i] is -1.

Example 1:

Input: n = 3, edges = [[0,1,2],[1,2,1],[0,2,4]], disappear = [1,1,5]

Output: [0,-1,4]

Explanation:

We are starting our journey from node 0, and our goal is to find the minimum time required to reach each node before it disappears.

For node 0, we don't need any time as it is our starting point.
For node 1, we need at least 2 units of time to traverse edges[0]. Unfortunately, it disappears at that moment, so we won't be able to visit it.
For node 2, we need at least 4 units of time to traverse edges[2].

Example 2:

Input: n = 3, edges = [[0,1,2],[1,2,1],[0,2,4]], disappear = [1,3,5]

Output: [0,2,3]

Explanation:

We are starting our journey from node 0, and our goal is to find the minimum time required to reach each node before it disappears.

For node 0, we don't need any time as it is the starting point.
For node 1, we need at least 2 units of time to traverse edges[0].
For node 2, we need at least 3 units of time to traverse edges[0] and edges[1].

Example 3:

Input: n = 2, edges = [[0,1,1]], disappear = [1,1]

Output: [0,-1]

Explanation:

Exactly when we reach node 1, it disappears.

Constraints:

1 <= n <= 5 * 10⁴
0 <= edges.length <= 10⁵
edges[i] == [u_i, v_i, length_i]
0 <= u_i, v_i <= n - 1
1 <= length_i <= 10⁵
disappear.length == n
1 <= disappear[i] <= 10⁵

Step 01

Brute Force Baseline

Problem summary: There is an undirected graph of n nodes. You are given a 2D array edges, where edges[i] = [ui, vi, lengthi] describes an edge between node ui and node vi with a traversal time of lengthi units. Additionally, you are given an array disappear, where disappear[i] denotes the time when the node i disappears from the graph and you won't be able to visit it. Note that the graph might be disconnected and might contain multiple edges. Return the array answer, with answer[i] denoting the minimum units of time required to reach node i from node 0. If node i is unreachable from node 0 then answer[i] is -1.

Baseline thinking

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

Pattern signal: Array

Example 1

3
[[0,1,2],[1,2,1],[0,2,4]]
[1,1,5]

Example 2

3
[[0,1,2],[1,2,1],[0,2,4]]
[1,3,5]

Example 3

2
[[0,1,1]]
[1,1]

Core Insight

What unlocks the optimal approach

Use Dijkstra’s algorithm, but only visit nodes if you can reach them before disappearance.

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 #3112: Minimum Time to Visit Disappearing Nodes
class Solution {
    public int[] minimumTime(int n, int[][] edges, int[] disappear) {
        List<int[]>[] g = new List[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (var e : edges) {
            int u = e[0], v = e[1], w = e[2];
            g[u].add(new int[] {v, w});
            g[v].add(new int[] {u, w});
        }
        int[] dist = new int[n];
        Arrays.fill(dist, 1 << 30);
        dist[0] = 0;
        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[0] - b[0]);
        pq.offer(new int[] {0, 0});
        while (!pq.isEmpty()) {
            var e = pq.poll();
            int du = e[0], u = e[1];
            if (du > dist[u]) {
                continue;
            }
            for (var nxt : g[u]) {
                int v = nxt[0], w = nxt[1];
                if (dist[v] > dist[u] + w && dist[u] + w < disappear[v]) {
                    dist[v] = dist[u] + w;
                    pq.offer(new int[] {dist[v], v});
                }
            }
        }
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            ans[i] = dist[i] < disappear[i] ? dist[i] : -1;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3112: Minimum Time to Visit Disappearing Nodes
func minimumTime(n int, edges [][]int, disappear []int) []int {
	g := make([][]pair, n)
	for _, e := range edges {
		u, v, w := e[0], e[1], e[2]
		g[u] = append(g[u], pair{v, w})
		g[v] = append(g[v], pair{u, w})
	}

	dist := make([]int, n)
	for i := range dist {
		dist[i] = 1 << 30
	}
	dist[0] = 0

	pq := hp{{0, 0}}

	for len(pq) > 0 {
		du, u := pq[0].dis, pq[0].u
		heap.Pop(&pq)

		if du > dist[u] {
			continue
		}

		for _, nxt := range g[u] {
			v, w := nxt.dis, nxt.u
			if dist[v] > dist[u]+w && dist[u]+w < disappear[v] {
				dist[v] = dist[u] + w
				heap.Push(&pq, pair{dist[v], v})
			}
		}
	}

	ans := make([]int, n)
	for i := 0; i < n; i++ {
		if dist[i] < disappear[i] {
			ans[i] = dist[i]
		} else {
			ans[i] = -1
		}
	}

	return ans
}

type pair struct{ dis, u int }
type hp []pair

func (h hp) Len() int           { return len(h) }
func (h hp) Less(i, j int) bool { return h[i].dis < h[j].dis }
func (h hp) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
func (h *hp) Push(v any)        { *h = append(*h, v.(pair)) }
func (h *hp) Pop() any          { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }

# Accepted solution for LeetCode #3112: Minimum Time to Visit Disappearing Nodes
class Solution:
    def minimumTime(
        self, n: int, edges: List[List[int]], disappear: List[int]
    ) -> List[int]:
        g = defaultdict(list)
        for u, v, w in edges:
            g[u].append((v, w))
            g[v].append((u, w))
        dist = [inf] * n
        dist[0] = 0
        pq = [(0, 0)]
        while pq:
            du, u = heappop(pq)
            if du > dist[u]:
                continue
            for v, w in g[u]:
                if dist[v] > dist[u] + w and dist[u] + w < disappear[v]:
                    dist[v] = dist[u] + w
                    heappush(pq, (dist[v], v))
        return [a if a < b else -1 for a, b in zip(dist, disappear)]

// Accepted solution for LeetCode #3112: Minimum Time to Visit Disappearing Nodes
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3112: Minimum Time to Visit Disappearing Nodes
// class Solution {
//     public int[] minimumTime(int n, int[][] edges, int[] disappear) {
//         List<int[]>[] g = new List[n];
//         Arrays.setAll(g, k -> new ArrayList<>());
//         for (var e : edges) {
//             int u = e[0], v = e[1], w = e[2];
//             g[u].add(new int[] {v, w});
//             g[v].add(new int[] {u, w});
//         }
//         int[] dist = new int[n];
//         Arrays.fill(dist, 1 << 30);
//         dist[0] = 0;
//         PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[0] - b[0]);
//         pq.offer(new int[] {0, 0});
//         while (!pq.isEmpty()) {
//             var e = pq.poll();
//             int du = e[0], u = e[1];
//             if (du > dist[u]) {
//                 continue;
//             }
//             for (var nxt : g[u]) {
//                 int v = nxt[0], w = nxt[1];
//                 if (dist[v] > dist[u] + w && dist[u] + w < disappear[v]) {
//                     dist[v] = dist[u] + w;
//                     pq.offer(new int[] {dist[v], v});
//                 }
//             }
//         }
//         int[] ans = new int[n];
//         for (int i = 0; i < n; ++i) {
//             ans[i] = dist[i] < disappear[i] ? dist[i] : -1;
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3112: Minimum Time to Visit Disappearing Nodes
function minimumTime(n: number, edges: number[][], disappear: number[]): number[] {
    const g: [number, number][][] = Array.from({ length: n }, () => []);
    for (const [u, v, w] of edges) {
        g[u].push([v, w]);
        g[v].push([u, w]);
    }
    const dist = Array.from({ length: n }, () => Infinity);
    dist[0] = 0;
    const pq = new PriorityQueue({
        compare: (a, b) => (a[0] === b[0] ? a[1] - b[1] : a[0] - b[0]),
    });
    pq.enqueue([0, 0]);
    while (pq.size() > 0) {
        const [du, u] = pq.dequeue()!;
        if (du > dist[u]) {
            continue;
        }
        for (const [v, w] of g[u]) {
            if (dist[v] > dist[u] + w && dist[u] + w < disappear[v]) {
                dist[v] = dist[u] + w;
                pq.enqueue([dist[v], v]);
            }
        }
    }
    return dist.map((a, i) => (a < disappear[i] ? a : -1));
}

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(m × log m)

Space

O(m)

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.

Minimum Time to Visit Disappearing Nodes

Problem Statement

Roadmap

Brute Force Baseline

Baseline thinking

Example 1

Example 2

Example 3

Related Problems

Core Insight

What unlocks the optimal approach

Algorithm Walkthrough

Iteration Checklist

Edge Cases

Full Annotated Code

Interactive Study Demo

Complexity Analysis

Approach Breakdown

Common Mistakes

Off-by-one on range boundaries