# 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;
    }
};
```