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.
A 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 <= 200n == isConnected.lengthn == isConnected[i].lengthisConnected[i][j] is 1 or 0.isConnected[i][i] == 1isConnected[i][j] == isConnected[j][i]
class Solution {
public:
void provinces(unordered_map>& mp, vector &visited, int curr)
{
visited[curr]=true;
for(auto i:mp[curr])
{
if(!visited[i])
{
provinces(mp,visited,i);
}
}
}
int findCircleNum(vector>& isConnected) {
unordered_map> mp;
int n=isConnected.size();
for(int i=0; i visited(n,false);
int count=0;
for(int i=0; i
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