# 28. 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:**![](https://assets.leetcode.com/uploads/2020/01/16/find_the_city_01.png)

```
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:**![](https://assets.leetcode.com/uploads/2020/01/16/find_the_city_02.png)

```
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.
```

## Solution: (Dijkstra’s Algorithm on all source nodes)

```cpp
class Solution
{
public:
    int res = INT_MAX;
    int prevCount = INT_MAX;

    void findSortestPath(int src, vector<pair<int, int>> v[], int n, int threshold)
    {
        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;

        vector<int> dist(n, INT_MAX);
        vector<bool> vs(n, false);

        dist[src] = 0;
        pq.push({0, src});

        while (!pq.empty())
        {

            int weight = pq.top().first;
            int node = pq.top().second;
            pq.pop();

            if (vs[node])
            {
                continue;
            }

            vs[node] = 1;

            for (int i = 0; i < v[node].size(); i++)
            {
                int cn = v[node][i].first;
                int cw = v[node][i].second;

                if (cw + weight < dist[cn])
                {
                    dist[cn] = cw + weight;
                    pq.push({dist[cn], cn});
                }
            }
        }
        int count = 0;
        for (int i = 0; i < n; i++)
        {
            if (i == src)
            {
                continue;
            }

            if (dist[i] <= threshold)
            {
                count++;
            }
        }
        
        if (count <= prevCount)
        {
            prevCount = count;
            res = src;
        }

        return;
    }

    int findTheCity(int n, vector<vector<int>> &edges, int distanceThreshold)
    {

        vector<pair<int, int>> v[n];

        for (int i = 0; i < edges.size(); i++)
        {
            int a = edges[i][0];
            int b = edges[i][1];
            int w = edges[i][2];
            v[a].push_back({b, w});
            v[b].push_back({a, w});
        }

        for (int i = 0; i < n; i++)
        {
            findSortestPath(i, v, n, distanceThreshold);
        }

        return res;
    }
};
```

## Solution: (Floyd Warshall)

```cpp
class Solution
{
public:
    void findSortestPath(vector<vector<int>> &v, int n)
    {
        for (int k = 0; k < n; k++)
        {
            for (int i = 0; i < n; i++)
            {
                for (int j = 0; j < n; j++)
                {
                    v[i][j] = min(v[i][j], v[i][k] + v[k][j]);
                }
            }
        }
    }

    int findTheCity(int n, vector<vector<int>> &edges, int distanceThreshold)
    {

        vector<vector<int>> v(n, vector<int>(n, 100000));

        for (int i = 0; i < n; i++)
        {
            v[i][i] = 0;
        }

        for (int i = 0; i < edges.size(); i++)
        {
            int a = edges[i][0];
            int b = edges[i][1];
            int w = edges[i][2];
            v[a][b] = w;
            v[b][a] = w;
        }

        findSortestPath(v, n);
        int prevCount = INT_MAX;
        int res = -1;

        for (int i = 0; i < n; i++)
        {
            int count = 0;
            for (int j = 0; j < n; j++)
            {
                if (v[i][j] <= distanceThreshold)
                {
                    count++;
                }
            }

            if (count <= prevCount)
            {
                prevCount = count;
                res = i;
            }
        }

        return res;
    }
};
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://soumyajit4419.gitbook.io/ds-algo/graph/28.-find-the-city-with-the-smallest-number-of-neighbors-at-a-threshold-distance.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.