How do you approach Number of Provinces?

Treat the matrix as an adjacency matrix and union connected pairs.

What are common mistakes on Number of Provinces?

Building the adjacency list in one direction on an undirected graph. Assuming the graph is connected. Most inputs allow several components, and one traversal from node 0 misses the rest. Ignoring self-loops and parallel edges when the constraints allow them.

gitGood.dev

← Back to problems

Number of Provinces

MediumAlgorithms

Description

There are `n` cities. Some are connected directly or indirectly. A **province** is a group of directly or indirectly connected cities. You are given an `n x n` matrix `isConnected` where `isConnected[i][j] = 1` if city `i` and city `j` are directly connected. Return the total number of provinces.

Example 1

Input: isConnected = [[1,1,0],[1,1,0],[0,0,1]]

Output: 2

Explanation: {0,1} and {2}.

Example 2

Input: isConnected = [[1,0,0],[0,1,0],[0,0,1]]

Output: 3

Explanation: No city is connected to another.

Constraints

- 1 <= n <= 200 - isConnected[i][i] == 1 - isConnected[i][j] == isConnected[j][i]

Number of Provinces: approach, complexity and pitfalls

A medium graphs problem, graded against 4 test cases (2 of them hidden).

The pattern: Graphs

Nodes and edges - traversal, connectivity, cycles, and shortest paths.

Reach for it when you see: Explicit edges, a grid treated as a graph, or dependencies between items.

More Graphsproblems →

If you are stuck

Show the 2-step hint ladder

Treat the matrix as an adjacency matrix and union connected pairs.
Only the upper triangle (j > i) needs scanning since the matrix is symmetric.

The optimal approach

Show the approach: The approach

Union-Find over the adjacency matrix. Union every pair `(i, j)` with `isConnected[i][j] == 1`, starting from `count = n` and decrementing on each merge. The remaining count is the number of provinces.

Time: O(n²·α(n)). Space: O(n).

The full reference solution in every supported language stays in the editor above - reveal it there once you have had a real attempt.

What the tests check

Read off this problem's own test suite, so these are the cases a submission actually has to survive.

Single element. A one-element case is included. Two-pointer and window solutions frequently return the wrong thing when the two indices start out equal.
Duplicate values. Repeated values appear in the input, so any logic that assumes distinct elements - or that skips over equal neighbours - is exercised.
Zero. Zero appears in the input, which is the value most likely to break a truthiness check used in place of an explicit comparison.

Common Graphs mistakes

These apply to the pattern as a whole, not just this problem.

Building the adjacency list in one direction on an undirected graph.
Assuming the graph is connected. Most inputs allow several components, and one traversal from node 0 misses the rest.
Ignoring self-loops and parallel edges when the constraints allow them.