Find the City with Fewest Reachable Neighbours
A medium Graph problem included in Striver A2Z. Below: the roles whose interviews prioritise this topic, and how to practise it.
- Topic
- Graph
- Sheets
- 1
- Core for
- 11 roles
- Platform
- LeetCode
The problem
Given an undirected weighted graph with n nodes labeled from 0 to n-1, a list of edges [u, v, weight], and a distance threshold, find the node with the smallest number of neighbors reachable within the distance threshold. If multiple nodes qualify, return the one with the largest label.
Example 1
- Input
- n = 4, edges = [[0,1,3],[1,2,1],[1,3,4],[2,3,1]], distanceThreshold = 4
- Output
- 3
- Why
- Node 3 can reach 2 neighbors within distance 4. Node 0 also has 2 but node 3 has the larger label.
Example 2
- Input
- n = 5, edges = [[0,1,2],[0,4,8],[1,2,3],[1,4,2],[2,3,1],[3,4,7]], distanceThreshold = 2
- Output
- 0
- Why
- Node 0 can reach only node 1 within distance 2. Other nodes have more reachable neighbors.
Constraints
- 2 <= n <= 100
- 0 <= edges.length <= n * (n - 1) / 2
- 1 <= distanceThreshold <= 10^4
How to think about it
Updated 2026-09-09Because n is at most 100, finding all-pairs shortest paths via Floyd-Warshall is compact and fast. After filling the distance matrix, count how many other cities each city i can reach with distance <= distanceThreshold. Walk cities from 0 to n - 1: if a city's reachable count is less than or equal to the current minimum, it becomes the new best candidate (the equality condition naturally gives preference to the larger index).
Approaches, worst first
Floyd-Warshall all-pairs matrix
time O(n^3) · space O(n^2)
Build an n x n matrix initialized to infinity and 0 on the diagonal. Relax via triple loop over intermediate k. For each city i, count j != i where `dist[i][j] <= distanceThreshold`. Select the city with minimal count, breaking ties by larger city index.
Repeated Dijkstra from every nodeWrite this one
time O(n * E log n) · space O(n + E)
Run Dijkstra from each city i = 0 to n - 1. Count nodes whose shortest path is within threshold. Particularly useful if the graph is sparse, running in O(n * E log n).
Where people lose marks · 3
- Tie-breaking rule: if two cities have the same minimum number of reachable neighbors, choose the one with the LARGER index; using strict `<` keeps the smaller index instead.
- Do not count the city itself as one of its reachable neighbors (`j !== i`).
- Edges are undirected; failing to add both directions `(u, v)` and `(v, u)` corrupts symmetric distances.
The theory behind it
Graph — the ground this problem stands on. All Graph problems
What Graph is
A graph is a network of individual points, called vertices or nodes, connected by lines called edges. Think of a subway transit map, an electrical circuit, or a web of social friends. Unlike a tree, a graph has no designated top node and no parent-child hierarchy. Connections can run one-way or both ways, and paths can loop back on themselves to form closed cycles.
When to reach for it
Reach for graph algorithms when inputs describe relationships, networks, flights between cities, course prerequisites, or clone networks. Signals include finding the shortest route across unweighted connections, ordering tasks that depend on earlier tasks, counting isolated clusters, or checking whether a path contains an infinite loop. Whenever problems present pairs of related entities and ask for reachability, distances, or dependencies, graph representations apply.
How the pattern works
First convert edge lists into an adjacency list, mapping each node to an array of its neighbors. Choose your exploration strategy based on the goal: use a queue and breadth-first search to find the shortest path in unweighted networks, or use recursion and depth-first search to explore full paths and detect cycles. Because graphs can have loops, always track visited nodes in a set or boolean array. Add nodes to the visited set at the moment they enter the queue so they are never visited twice.
What each operation costs
| Operation | Time |
|---|---|
| visit all nodes and edges via search | O(v + e) |
| topological sort using in-degree counts | O(v + e) |
| shortest path using dijkstra with a min-heap | O((v + e) log v) |
What usually goes wrong with Graph
- Adding a node to the visited set when popping from the queue instead of when pushing, which lets neighboring nodes enqueue duplicate entries and wastes memory.
- Failing to check for cycles in directed graphs when finding prerequisite orders, causing topological sort routines to hang or return incomplete lists.
- Assuming an input graph is fully connected and scanning from only a single starting node, missing disconnected islands and isolated components.
Which roles need this problem
Graph is a core topic for these 11 roles — if you're targeting one of them, this problem is early in your path, not optional.
Secondary for 6 more roles, including Performance Engineer, Search Engineer, Information Retrieval Engineer.
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