1334. Find the City With the Smallest Number of Neighbors at a Threshold Distance LeetCode Solution
In this guide, you will get 1334. Find the City With the Smallest Number of Neighbors at a Threshold Distance LeetCode Solution with the best time and space complexity. The solution to Find the City With the Smallest Number of Neighbors at a Threshold Distance problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Table of Contents
- Problem Statement
- Complexity Analysis
- Find the City With the Smallest Number of Neighbors at a Threshold Distance solution in C++
- Find the City With the Smallest Number of Neighbors at a Threshold Distance solution in Java
- Find the City With the Smallest Number of Neighbors at a Threshold Distance solution in Python
- Additional Resources
Problem Statement of Find the City With the Smallest Number of Neighbors at a Threshold Distance
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.
Return the city with the smallest number of cities that are reachable through some path and whose distance is at most distanceThreshold, If there are multiple such cities, return the city with the greatest number.
Notice that the distance of a path connecting cities i and j is equal to the sum of the edges’ weights along that path.
Example 1:
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.
Example 2:
Input: n = 5, edges = [[0,1,2],[0,4,8],[1,2,3],[1,4,2],[2,3,1],[3,4,1]], distanceThreshold = 2
Output: 0
Explanation: The figure above describes the graph.
The neighboring cities at a distanceThreshold = 2 for each city are:
City 0 -> [City 1]
City 1 -> [City 0, City 4]
City 2 -> [City 3, City 4]
City 3 -> [City 2, City 4]
City 4 -> [City 1, City 2, City 3]
The city 0 has 1 neighboring city at a distanceThreshold = 2.
Constraints:
2 <= n <= 100
1 <= edges.length <= n * (n – 1) / 2
edges[i].length == 3
0 <= fromi < toi < n
1 <= weighti, distanceThreshold <= 10^4
All pairs (fromi, toi) are distinct.
Complexity Analysis
- Time Complexity:
- Space Complexity:
1334. Find the City With the Smallest Number of Neighbors at a Threshold Distance LeetCode Solution in C++
class Solution {
public:
int findTheCity(int n, vector<vector<int>>& edges, int distanceThreshold) {
int ans = -1;
int minCitiesCount = n;
const vector<vector<int>> dist = floydWarshall(n, edges, distanceThreshold);
for (int i = 0; i < n; ++i) {
int citiesCount = 0;
for (int j = 0; j < n; ++j)
if (dist[i][j] <= distanceThreshold)
++citiesCount;
if (citiesCount <= minCitiesCount) {
ans = i;
minCitiesCount = citiesCount;
}
}
return ans;
}
private:
vector<vector<int>> floydWarshall(int n, const vector<vector<int>>& edges,
int distanceThreshold) {
vector<vector<int>> dist(n, vector<int>(n, distanceThreshold + 1));
for (int i = 0; i < n; ++i)
dist[i][i] = 0;
for (const vector<int>& edge : edges) {
const int u = edge[0];
const int v = edge[1];
const int w = edge[2];
dist[u][v] = w;
dist[v][u] = w;
}
for (int k = 0; k < n; ++k)
for (int i = 0; i < n; ++i)
for (int j = 0; j < n; ++j)
dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
return dist;
}
};
/* code provided by PROGIEZ */1334. Find the City With the Smallest Number of Neighbors at a Threshold Distance LeetCode Solution in Java
class Solution {
public int findTheCity(int n, int[][] edges, int distanceThreshold) {
int ans = -1;
int minCitiesCount = n;
int[][] dist = floydWarshall(n, edges, distanceThreshold);
for (int i = 0; i < n; ++i) {
int citiesCount = 0;
for (int j = 0; j < n; ++j)
if (dist[i][j] <= distanceThreshold)
++citiesCount;
if (citiesCount <= minCitiesCount) {
ans = i;
minCitiesCount = citiesCount;
}
}
return ans;
}
private int[][] floydWarshall(int n, int[][] edges, int distanceThreshold) {
int[][] dist = new int[n][n];
Arrays.stream(dist).forEach(A -> Arrays.fill(A, distanceThreshold + 1));
for (int i = 0; i < n; ++i)
dist[i][i] = 0;
for (int[] edge : edges) {
final int u = edge[0];
final int v = edge[1];
final int w = edge[2];
dist[u][v] = w;
dist[v][u] = w;
}
for (int k = 0; k < n; ++k)
for (int i = 0; i < n; ++i)
for (int j = 0; j < n; ++j)
dist[i][j] = Math.min(dist[i][j], dist[i][k] + dist[k][j]);
return dist;
}
}
// code provided by PROGIEZ1334. Find the City With the Smallest Number of Neighbors at a Threshold Distance LeetCode Solution in Python
class Solution:
def findTheCity(
self,
n: int,
edges: list[list[int]],
distanceThreshold: int,
) -> int:
ans = -1
minCitiesCount = n
dist = self._floydWarshall(n, edges, distanceThreshold)
for i in range(n):
citiesCount = sum(dist[i][j] <= distanceThreshold for j in range(n))
if citiesCount <= minCitiesCount:
ans = i
minCitiesCount = citiesCount
return ans
def _floydWarshall(
self,
n: int,
edges: list[list[int]],
distanceThreshold: int,
) -> list[list[int]]:
dist = [[distanceThreshold + 1] * n for _ in range(n)]
for i in range(n):
dist[i][i] = 0
for u, v, w in edges:
dist[u][v] = w
dist[v][u] = w
for k in range(n):
for i in range(n):
for j in range(n):
dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j])
return dist
# code by PROGIEZAdditional Resources
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