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| 1 | +class Solution |
| 2 | +{ |
| 3 | + public: |
| 4 | + // Function to perform Depth First Search (DFS) and populate the stack with nodes in order of their finish time |
| 5 | + void DFS(int node, vector<int>& visited, vector<vector<int>>& adjacencyList, stack<int>& nodesOrder) { |
| 6 | + // Mark the current node as visited |
| 7 | + visited[node] = true; |
| 8 | + |
| 9 | + // Explore all unvisited neighbors of the current node |
| 10 | + for (auto neighbour : adjacencyList[node]) { |
| 11 | + if (!visited[neighbour]) |
| 12 | + DFS(neighbour, visited, adjacencyList, nodesOrder); |
| 13 | + } |
| 14 | + |
| 15 | + // After visiting all neighbors, push the current node onto the stack |
| 16 | + nodesOrder.push(node); |
| 17 | + } |
| 18 | + |
| 19 | + // Function to perform DFS on the transposed graph |
| 20 | + void reverseDFS(int node, vector<vector<int>>& adjacencyList, vector<int>& visited) { |
| 21 | + // Mark the current node as visited |
| 22 | + visited[node] = true; |
| 23 | + |
| 24 | + // Explore all unvisited neighbors of the current node in the transposed graph |
| 25 | + for (auto neighbour : adjacencyList[node]) { |
| 26 | + if (!visited[neighbour]) |
| 27 | + reverseDFS(neighbour, adjacencyList, visited); |
| 28 | + } |
| 29 | + } |
| 30 | + |
| 31 | + // Function to find the number of strongly connected components (SCCs) using Kosaraju's Algorithm |
| 32 | + int kosaraju(int V, vector<vector<int>>& adj) { |
| 33 | + // Step 1: Initialize visited vector and stack to store nodes in order of their finish time |
| 34 | + vector<int> visited(V, false); |
| 35 | + stack<int> nodesOrder; |
| 36 | + vector<vector<int>> transpose(V); // Adjacency list for the transposed graph |
| 37 | + int totalSCCs = 0; // Counter for SCCs |
| 38 | + |
| 39 | + // Step 2: Perform DFS on the original graph to populate the stack |
| 40 | + for (int i = 0; i < V; i++) { |
| 41 | + if (!visited[i]) |
| 42 | + DFS(i, visited, adj, nodesOrder); |
| 43 | + } |
| 44 | + |
| 45 | + // Step 3: Create the transposed graph |
| 46 | + for (int i = 0; i < V; i++) { |
| 47 | + for (auto neighbour : adj[i]) { |
| 48 | + // Reverse the direction of edges |
| 49 | + transpose[neighbour].push_back(i); |
| 50 | + } |
| 51 | + } |
| 52 | + |
| 53 | + // Step 4: Reset visited array for use in the second DFS |
| 54 | + fill(visited.begin(), visited.end(), false); |
| 55 | + |
| 56 | + // Step 5: Perform DFS on the transposed graph in the order of nodes in the stack |
| 57 | + while (!nodesOrder.empty()) { |
| 58 | + int node = nodesOrder.top(); |
| 59 | + nodesOrder.pop(); |
| 60 | + |
| 61 | + // If the node is not visited, it means we've found a new SCC |
| 62 | + if (!visited[node]) { |
| 63 | + totalSCCs++; // Increment SCC count |
| 64 | + reverseDFS(node, transpose, visited); // Explore all nodes in this SCC |
| 65 | + } |
| 66 | + } |
| 67 | + |
| 68 | + // Step 6: Return the total number of SCCs |
| 69 | + return totalSCCs; |
| 70 | + } |
| 71 | +}; |
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