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# Number of Provinces LeetCode Solution

## Problem – 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]`

### Number of Provinces LeetCode Solution in Java

``````public class Solution {
public void dfs(int[][] M, int[] visited, int i) {
for (int j = 0; j < M.length; j++) {
if (M[i][j] == 1 && visited[j] == 0) {
visited[j] = 1;
dfs(M, visited, j);
}
}
}
public int findCircleNum(int[][] M) {
int[] visited = new int[M.length];
int count = 0;
for (int i = 0; i < M.length; i++) {
if (visited[i] == 0) {
dfs(M, visited, i);
count++;
}
}
return count;
}
}
``````

### Number of Provinces LeetCode Solution in Python

``````def findCircleNum(self, A):
N = len(A)
seen = set()
def dfs(node):
if adj and nei not in seen:
dfs(nei)

ans = 0
for i in xrange(N):
if i not in seen:
dfs(i)
ans += 1
return ans
``````

### Number of Provinces LeetCode Solution in C++

``````class Solution {
public:
int findCircleNum(vector<vector<int>>& M) {
if (M.empty()) return 0;
int n = M.size();
vector<bool> visited(n, false);
int groups = 0;
for (int i = 0; i < visited.size(); i++) {
groups += !visited[i] ? dfs(i, M, visited), 1 : 0;
}
return groups;
}

private:
void dfs(int i, vector<vector<int>>& M, vector<bool>& visited) {
visited[i] = true;
for (int j = 0; j < visited.size(); j++) {
if (i != j && M[i][j] && !visited[j]) {
dfs(j, M, visited);
}
}
}
};
``````
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