### LeetCode Challenge #547. Number of Provinces

There are `n` cities. Some of them are connected, while some are not. If city `a` is connected directly with city `b`, and city `b` is connected directly with city `c`, then city `a` is connected indirectly with city `c`.

province is a group of directly or indirectly connected cities and no other cities outside of the group.

You are given an `n x n` matrix `isConnected` where `isConnected[i][j] = 1` if the `ith` city and the `jth` city are directly connected, and `isConnected[i][j] = 0` otherwise.

Return the total number of provinces.

Example 1:

```Input: isConnected = [[1,1,0],[1,1,0],[0,0,1]]
Output: 2
```

Example 2:

```Input: isConnected = [[1,0,0],[0,1,0],[0,0,1]]
Output: 3
```

Constraints:

• `1 <= n <= 200`
• `n == isConnected.length`
• `n == isConnected[i].length`
• `isConnected[i][j]` is `1` or `0`.
• `isConnected[i][i] == 1`
• `isConnected[i][j] == isConnected[j][i]`
##### Video Solution
###### C++ Solution
```				```
class Solution {
public:
void provinces(unordered_map<int,vector<int>>& mp, vector<bool> &visited, int curr)
{
visited[curr]=true;
for(auto i:mp[curr])
{
if(!visited[i])
{
provinces(mp,visited,i);
}
}
}
int findCircleNum(vector<vector<int>>& isConnected) {
unordered_map<int,vector<int>> mp;
int n=isConnected.size();

for(int i=0; i<n; i++)
{
for(int j=0; j<n; j++)
{
if(isConnected[i][j]==1 and i!=j)mp[i].push_back(j);
}
}

vector<bool> visited(n,false);
int count=0;

for(int i=0; i<n; i++)
{
if(!visited[i])
{
provinces(mp,visited,i);
count++;
}
}
return count;
}
};
```
```