Sure, here's a refined and simplified title focusing on the product keywords: EVORETRO Cartridge Protectors - Clear Plastic Display Case for 8-Bit Video Games (25 Boxes)

"Given a set of n nodes, where n > 0, and m directed edges, where m >= 0, the objective is to find the number of strongly connected components (SCCs) in the directed graph. A strongly connected component is a subset of nodes where each node in the subset is reachable from every other node in the subset.

The solution to this problem can be achieved using Kosaraju's algorithm, which consists of the following steps:

1. Perform a depth-first search (DFS) on the graph and record the finishing time of each node in a stack.
2. Transpose the graph, meaning that the direction of each edge is reversed.
3. Perform another DFS on the transposed graph, starting from the nodes in the order of decreasing finishing times (from the stack).
4. Each time a new node is visited during this second DFS, a new SCC is formed.

The time complexity of this algorithm is O(V + E), where V is the number of nodes and E is the number of edges in the graph.

Here's the Python code that implements Kosaraju's algorithm:

```python def kosaraju(graph): n = len(graph) visited = [False] * n stack = []

# Step 1: Perform DFS and record finishing times
for i in range(n):
    if not visited[i]:
        dfs1(graph, i, visited, stack)

# Step 2: Transpose the graph
transposed_graph = [[] for _ in range(n)]
for i in range(n):
    for neighbor in graph[i]:
        transposed_graph[neighbor].append(i)

# Step 3: Perform DFS on the transposed graph
visited = [False] * n
sccs = 0
while stack:
    node = stack.pop()
    if not visited[node]:
        dfs2(transposed_graph, node, visited)
        sccs += 1

return sccs

def dfs1(graph, node, visited, stack): visited[node] = True for neighbor in graph[node]: if not visited[neighbor]: dfs1(graph, neighbor, visited, stack) stack.append(node)

def dfs2(graph, node, visited): visited[node] = True for neighbor in graph[node]: if not visited[neighbor]: dfs2(graph, neighbor, visited)

Example usage

graph = [[1], [2], [0, 3], [2]] print(kosaraju(graph)) # Output: 2 ```

The key steps in Kosaraju's algorithm are: 1. Perform a depth-first search (DFS) on the graph and record the finishing time of each node in a stack. 2. Transpose the graph, meaning that the direction of each edge is reversed. 3. Perform another DFS on the transposed graph, starting from the nodes in the order of decreasing finishing times (from the stack). 4. Each time a new node is visited during this second DFS, a new SCC is formed.

The time complexity of this algorithm is O(V + E), where V is the number of nodes and E is the number of edges in the graph.

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