Provinces Solution

Process the adjacency matrix, union connected cities:

function findCircleNum(isConnected):
    n = isConnected.length
    uf = new UnionFind(n)

    for i from 0 to n-1:
        for j from i+1 to n-1:
            if isConnected[i][j] == 1:
                uf.union(i, j)

    return uf.count

Only check $j > i$ since the matrix is symmetric. This avoids processing each edge twice.

Alternative: DFS from each unvisited node, counting the number of DFS starts. Both approaches work; Union-Find is cleaner here.

Time: $O(n^2 \cdot \alpha(n)) \approx O(n^2)$. Space: $O(n)$.