Number of Provinces

Problem Statement:

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:

  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

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✅ Solution: Connected Components Counting (Find Number of Provinces / Connected Groups)

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✅ Code — DFS + Adjacency List Construction

// ✅ Find Number of Provinces Using DFS and Adjacency List Conversion

class Solution {
public:
    void dfs(int node, vector<bool>& visited, vector<vector<int>>& adj) {
        visited[node] = true;

        for (auto neighbor : adj[node]) {
            if (!visited[neighbor]) {
                dfs(neighbor, visited, adj);
            }
        }
    }

    int findCircleNum(vector<vector<int>>& isConnected) {
        int v = isConnected.size();
        vector<vector<int>> adj(v);

        // Convert adjacency matrix to adjacency list
        for (int i = 0; i < v; i++) {
            for (int j = 0; j < v; j++) {
                if (isConnected[i][j] == 1 && i != j) {
                    adj[i].push_back(j);
                    adj[j].push_back(i);
                }
            }
        }

        vector<bool> visited(v, false);
        int count = 0;

        for (int i = 0; i < v; i++) {
            if (!visited[i]) {
                count++;
                dfs(i, visited, adj);
            }
        }

        return count;
    }
};

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📝 How It Works

• Step 1: Convert adjacency matrix isConnected into an adjacency list adj.

  ✅ This reduces redundant checks during DFS traversal.

• Step 2: Initialize visited[] array.

• Step 3: For each unvisited node:

  • Increment count (new province).
  • Call DFS to visit all nodes in that component.

• Step 4: Return the total number of provinces found.

✅ This is equivalent to counting connected components in an undirected graph.

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🧩 Key Formula / Recurrence

• DFS Recurrence:

  dfs(node) → for each neighbor → if not visited → dfs(neighbor)

• Final Result:

  count = number of DFS calls from unvisited nodes.

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⏱️ Time & Space Complexity

Metric	Value
Time	O(V²)
Space	O(V²)

• V: Number of vertices (cities).
• Matrix conversion takes O(V²).
• DFS traversal covers all edges once.

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⚠️ Edge Cases

• All cities disconnected → count = V.
• All cities fully connected → count = 1.
• Self-loops in isConnected[i][i] ignored (i ≠ j check).

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💡 Other Approaches

Approach	Time Complexity	Space Complexity	Notes
Adjacency Matrix DFS	O(V²)	O(V)	Skip adjacency list step.
Union-Find / Disjoint Set	O(V² * α(V))	O(V)	Faster for dynamic components.

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🔁 Related Problems

• Number of Islands
• Connected Components in Undirected Graph
• Redundant Connection Detection
• Friendship Circle Counting

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