547. Number of Provinces
1. Description
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.
2. Example
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
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]
4. Solutions
Breadth-First Search
n = isConnected.size()
Time complexity: O($n^2$)
Space complexity: O(n)
// Breadth-First Search and Depth-First Search are swappable
class Solution {
public:
int findCircleNum(const vector<vector<int>> &isConnected) {
vector<bool> visited(isConnected.size());
int result = 0;
queue<int> current_province;
for (int i = 0; i < isConnected.size(); ++i) {
if (!visited[i]) {
++result;
visited[i] = true;
current_province.push(i);
while (!current_province.empty()) {
int index = current_province.front();
current_province.pop();
for (int j = 0; j < isConnected[index].size(); ++j) {
if (!visited[j] && isConnected[index][j] == 1) {
visited[j] = true;
current_province.push(j);
}
}
}
}
}
return result;
}
};