Leetcode #1334: Find the City With the Smallest Number of Neighbors at a Threshold Distance
In this guide, we solve Leetcode #1334 Find the City With the Smallest Number of Neighbors at a Threshold Distance in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
There are n cities numbered from 0 to n-1. Given the array edges where edges[i] = [fromi, toi, weighti] represents a bidirectional and weighted edge between cities fromi and toi, and given the integer distanceThreshold.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Graph, Dynamic Programming, Shortest Path
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: n = 4, edges = [[0,1,3],[1,2,1],[1,3,4],[2,3,1]], distanceThreshold = 4
Output: 3
Explanation: The figure above describes the graph.
The neighboring cities at a distanceThreshold = 4 for each city are:
City 0 -> [City 1, City 2]
City 1 -> [City 0, City 2, City 3]
City 2 -> [City 0, City 1, City 3]
City 3 -> [City 1, City 2]
Cities 0 and 3 have 2 neighboring cities at a distanceThreshold = 4, but we have to return city 3 since it has the greatest number.
Python Solution
class Solution:
def findTheCity(
self, n: int, edges: List[List[int]], distanceThreshold: int
) -> int:
def dijkstra(u: int) -> int:
dist = [inf] * n
dist[u] = 0
vis = [False] * n
for _ in range(n):
k = -1
for j in range(n):
if not vis[j] and (k == -1 or dist[k] > dist[j]):
k = j
vis[k] = True
for j in range(n):
# dist[j] = min(dist[j], dist[k] + g[k][j])
if dist[k] + g[k][j] < dist[j]:
dist[j] = dist[k] + g[k][j]
return sum(d <= distanceThreshold for d in dist)
g = [[inf] * n for _ in range(n)]
for f, t, w in edges:
g[f][t] = g[t][f] = w
ans, cnt = n, inf
for i in range(n - 1, -1, -1):
if (t := dijkstra(i)) < cnt:
cnt, ans = t, i
return ans
Complexity
The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.