Number of Provinces medium
Description
There are n cities. Some of them are connected, while some are not. If city a is connected to city b, and city b is connected to city c, then city a is indirectly connected to 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 × n matrix isConnected where isConnected[i][j] = 1 if the ith and jth cities are directly connected, and 0 otherwise.
Return the total number of provinces.
Examples
> Case 1:
isConnected = [[1,1,0],[1,1,0],[0,0,1]]
Output: 2
# {0, 1} and {2}.
> Case 2:
isConnected = [[1,0,0],[0,1,0],[0,0,1]]
Output: 3Constraints
1 <= n <= 200n == isConnected.length == isConnected[i].lengthisConnected[i][j]is1or0.isConnected[i][i] == 1isConnected[i][j] == isConnected[j][i]
State design
Pure component counting. For every off-diagonal 1, union the two indices. The final number of distinct roots is the answer — and the components counter we maintain in the DSU template gives it directly.
Only need to walk the upper triangle (since the matrix is symmetric) for a small constant-factor win.
Code
Analysis
- Time:
O(n² · α(n))— scan the matrix once, near-constant per union. - Space:
O(n)for the DSU arrays.
DFS solves this too, in the same complexity. DSU wins when you want the “components after each new edge” running count. For one-shot static counting, both are fine — pick the one you can write fastest.
Same skin
- Number of Connected Components in an Undirected Graph — same idea, edge list instead of matrix.
- Friend Circles — older name for this exact problem.
- Number of Operations to Make Network Connected — components and spare edges, both via DSU.