3112. Minimum Time to Visit Disappearing Nodes

Problem Statement

There exists an undirected tree with n nodes numbered 0 to n-1. You are given a 2D integer array edges of length n-1, where edges[i] = [ui, vi, lengthi] indicates that there is an undirected edge between nodes ui and vi with a traversal time of lengthi seconds.

You are also given a 0-indexed array disappear, where disappear[i] is the time when node i disappears.

Return an array answer of length n, where answer[i] is the minimum time in seconds it takes to reach node i from node 0 before it disappears. If it’s impossible to reach node i before it disappears, return -1.

Examples

Example 1:

Input: n = 3, edges = [[0,1,2],[1,2,1],[0,2,4]], disappear = [1,1,5]
Output: [0,-1,4]
Explanation:
- Node 0: We start at node 0 at time 0, so answer[0] = 0.
- Node 1: We can reach node 1 at time 2, but disappear[1] = 1, so we cannot reach it in time. answer[1] = -1.
- Node 2: We can reach node 2 via path 0 -> 1 -> 2 at time 3, or directly via 0 -> 2 at time 4. Since disappear[2] = 5, we can reach it. The minimum time is 4, so answer[2] = 4.

Example 2:

Input: n = 3, edges = [[0,1,2],[1,2,1],[0,2,4]], disappear = [1,3,5]
Output: [0,2,3]
Explanation:
- Node 0: We start at node 0 at time 0, so answer[0] = 0.
- Node 1: We can reach node 1 at time 2, and disappear[1] = 3, so we can reach it. answer[1] = 2.
- Node 2: We can reach node 2 via path 0 -> 1 -> 2 at time 3, and disappear[2] = 5, so we can reach it. answer[2] = 3.

Example 3:

Input: n = 2, edges = [[0,1,1]], disappear = [1,1]
Output: [0,-1]
Explanation:
- Node 0: We start at node 0 at time 0, so answer[0] = 0.
- Node 1: We can reach node 1 at time 1, but disappear[1] = 1, so we cannot reach it in time (must arrive strictly before disappear time). answer[1] = -1.

Constraints

  • 2 <= n <= 5 * 10^4
  • edges.length == n - 1
  • edges[i].length == 3
  • 0 <= ui < vi < n
  • 1 <= lengthi <= 10^5
  • disappear.length == n
  • 1 <= disappear[i] <= 10^5

Clarification Questions

Before diving into the solution, here are 5 important clarifications and assumptions to discuss during an interview:

  1. Disappear time interpretation: What does “disappear” mean? (Assumption: A node can only be visited if we arrive strictly before its disappear time. If we arrive at time t and disappear[i] = t, we cannot visit it)

  2. Starting node: Can node 0 disappear immediately? (Assumption: If disappear[0] == 0, we cannot start, so all nodes are unreachable. Return array of all -1)

  3. Graph type: Is this always a tree? (Assumption: Yes, edges.length == n - 1 guarantees it’s a tree, meaning there’s exactly one path between any two nodes)

  4. Edge weights: Are edge weights always positive? (Assumption: Yes, 1 <= lengthi <= 10^5, so we can use Dijkstra’s algorithm)

  5. Multiple paths: Since it’s a tree, is there only one path between nodes? (Assumption: Yes, but we still need to find the minimum time, which Dijkstra handles efficiently)

Interview Deduction Process (20 minutes)

Step 1: Brute-Force Approach (5 minutes)

Use BFS/DFS to explore all paths from node 0, keeping track of minimum time to reach each node while checking disappear constraints. However, since it’s a tree, there’s only one path between nodes, so this is actually O(n) but doesn’t handle the time constraint optimally.

Step 2: Semi-Optimized Approach (7 minutes)

Use BFS level-by-level, but this doesn’t work well for weighted graphs. We need to consider edge weights, not just hop count.

Step 3: Optimized Solution (8 minutes)

Use Dijkstra’s algorithm with a modification: when relaxing an edge (u, v), only update dist[v] if dist[u] + weight < disappear[v]. This ensures we only consider paths that arrive before nodes disappear.

Solution Approach

This is a shortest path problem with time constraints. We need to find the minimum time to reach each node from node 0, but nodes disappear at specific times.

Key Insights:

  1. Dijkstra’s Algorithm: Perfect choice since all edge weights are positive
  2. Disappear Constraint: Check dist[u] + weight < disappear[v] when relaxing edges
  3. Early Termination: If we pop a node with dist[u] >= disappear[u], skip processing its neighbors
  4. Return Format: Use -1 to indicate unreachable nodes or nodes that disappear before we can reach them

Algorithm:

  1. Build adjacency list from edges (undirected graph)
  2. Initialize dist[0] = 0, all others as -1 (or INF)
  3. Use priority queue to process nodes in order of minimum distance
  4. For each popped node u:
    • If dist[u] >= disappear[u], skip (cannot visit)
    • For each neighbor v:
      • Calculate newDist = dist[u] + weight
      • If newDist < disappear[v] and newDist < dist[v] (or dist[v] == -1), update and push

Solution Code

Solution 1: Using long long with LLONG_MAX

class Solution:
def minimumTime(self, n, edges, disappear):
    if disappear[0] == 0) return list[int>(n, -1:
    list[list[pair<int, int>>> adjs(n)
    for e in edges:
        u = e[0], v = e[1], d = e[2]
        adjs[u].emplace_back(v, d)
        adjs[v].emplace_back(u, d)
    list[long long> dist(n, LLONG_MAX)
    heapq[pair<long long, int>, list[pair<long long, int>>, greater<>> pq
    dist[0] = 0
    pq.emplace(:0, 0)
    while not not pq:
        [t, u] = pq.top() pq.pop()
        if(t > dist[u]) continue
        if(t >= disappear[u]) continue
        for([v, w]: adjs[u]) :
        long long nt = t + w
        if dist[v] > nt  and  nt < disappear[v]:
            dist[v] = nt
            pq.emplace(:dist[v], v)
list[int> rtn(n, -1)
for(i = 0 i < n i += 1) :
if(dist[i] != LLONG_MAX) rtn[i] = dist[i]
return rtn

Key Points:

  • Uses long long to handle large distances
  • Checks t >= disappear[u] after popping to skip nodes that already disappeared
  • Checks nt < disappear[v] when relaxing edges
  • Converts LLONG_MAX to -1 in final result

Solution 2: Using int with -1 (Cleaner)

class Solution:
def minimumTime(self, n, edges, disappear):
    list[list[pair<int, int>>> adj(n)
    for e in edges:
        u = e[0], v = e[1], w = e[2]
        adj[u].emplace_back(v, w)
        adj[v].emplace_back(u, w)
    list[int> dis(n, -1)
    dis[0] = 0
    heapq[pair<int, int>, list[pair<int, int>>, greater<>> pq
    pq.emplace(0, 0)
    while not not pq:
        [du, u] = pq.top()
        pq.pop()
        if(dis[u] != -1  and  du > dis[u]) continue
        for([v, w]: adj[u]) :
        nd = du + w
        if nd < disappear[v] * and  (dis[v] == -1  or  nd < dis[v]):
            dis[v] = nd
            pq.emplace(nd, v)
return dis

Key Points:

  • Uses int since constraints guarantee values fit in int
  • Uses -1 directly to represent unreachable nodes
  • Checks nd < disappear[v] when relaxing edges
  • Simpler and more readable

Complexity Analysis

Time Complexity: O((V + E) log V)

  • Building adjacency list: O(E)
  • Dijkstra’s algorithm: O((V + E) log V) with priority queue
  • Total: O((V + E) log V)

Space Complexity: O(V + E)

  • Adjacency list: O(E)
  • Distance array: O(V)
  • Priority queue: O(V)
  • Total: O(V + E)

Edge Cases

  1. Node 0 disappears immediately: If disappear[0] == 0, return all -1
  2. Node disappears exactly when we arrive: If arrival_time == disappear[i], return -1 (must arrive strictly before)
  3. Unreachable nodes: Nodes with no path from node 0 return -1
  4. Large edge weights: Edge weights can be up to 10^5, but total path length is bounded by disappear times

Template Reference

See Graph Templates: Dijkstra for the standard Dijkstra template and variant with node disappearance constraints.