Problem Description

In this LeetCode problem, we are given a directed graph that has n nodes numbered from 0 to n - 1. The interesting property of this graph is that it would form a tree if the edges were bi-directional, meaning that there are no cycles and the graph is connected when considering the edges as undirected. However, as the edges are actually directed, not all nodes may be reachable from a given node.

We are provided with a 2D integer array edges, which represents the directed edges of the graph, where each element edges[i] = [u_i, v_i] denotes an edge from node u_i to node v_i.

An edge reversal is an operation that changes the direction of a directed edge, so an edge from u to v would be reversed to point from v to u instead.

The task is to calculate, for every node i, the minimum number of edge reversals that must be performed so that starting from node i, it would be possible to reach every other node in the graph. The solution should be returned as an integer array answer, where answer[i] gives the minimum number of edge reversals starting from node i.

Flowchart Walkthrough

Let's analyze the problem using the algorithm Flowchart. Here's a structured approach:

Is it a graph?

• Yes: The problem's description involves nodes and edges, which constitute a graph.

Is it a tree?

• No: Since we talk about edge reversals to make every node reachable, it indicates the presence of directed edges, hence not forming a strict tree since all nodes might not be reachable initially from a single root.

Is the problem related to directed acyclic graphs (DAGs)?

• No: The problem discusses edge reversals which directly implies the presence of cycles, otherwise no reversal would be needed if it were a DAG since DAGs inherently manage reachability without requiring reversals.

Does the problem involve connectivity?

• Yes: The main goal is to alter the graph's structure (by reversing edges) so that every node becomes reachable, explicitly concentrating on connectivity.

Is the graph weighted?

• No: The problem statement doesn’t specify weights on edges but focuses more on the structural properties (i.e., reachability by reversing edges).

Conclusion: According to the flowchart, for an unweighted connectivity problem that is not a DAG and requires a reachability solution in a potentially cyclic graph, DFS (Depth-First Search) would be a suitable pattern. DFS is apt for exploring the graph thoroughly, marking reachable nodes, and would help in tracking the paths for potential edge reversals to achieve full connectivity.

Intuition

The solution to this problem involves two main steps. First, we construct a graph representation that can help us determine the number of reversals needed to make all nodes accessible from a starting node. This is done by capturing not just the connection between nodes, but also the "directionality" with a value that indicates whether an edge needs to be reversed or not (signified by 1 for a properly directed edge, and -1 for an edge that requires reversal).

Once the graph is constructed, we commence a Depth First Search (DFS) to traverse the tree starting from an arbitrary root node (in this case, chosen to be node 0). During this traversal, we accumulate the number of reversals needed - which is when we encounter an edge where value k is negative.

After the initial pass, we do a second DFS to calculate the minimum number of reversals for all nodes based on the previously obtained information stored in ans[0]. This second DFS leverages the property that a reversal affects the reachability of all the nodes in the subtree of the reversed edge. Thus, by adjusting ans[i] for each node i by the direction of the incoming edge k, we propagate the required count of edge reversals throughout the tree.

This approach intuitively understands that in order to make all nodes reachable from one point, the direction of some edges need to align appropriately - which is essentially forming an outward spanning tree from the starting node, with edges oriented away from it.

Learn more about Depth-First Search, Breadth-First Search, Graph and Dynamic Programming patterns.

Solution Approach

The implemented solution for the problem uses the Depth First Search (DFS) algorithm twice on the graph. The graph is given a special structure where every edge is represented by a tuple containing the node it leads to and an integer indicating whether the edge points towards the node (1) or it needs to be reversed (-1).

Here are the detailed steps of the solution implementation:

1. The graph g is initialized as a list of lists, to hold the adjacency list representation of the graph. Each list within g corresponds to a node and contains tuples of its adjacent nodes and the directionality of the edges.

2. The dfs function is defined to traverse the graph starting from the root node 0. During the traversal, if a reversal is needed (indicated by a negative value), the number of necessary edge reversals ans[0] is incremented. The dfs function is called with the parameters i (the current node being visited) and fa (the node from which the current node was reached).

   • The function iterates over all the nodes j adjacent to the current node i.
   • If j is not the parent node fa, it inspects the direction k of the edge between i and j.
   • If k is negative, it means the edge is directed towards i and not away from it, and one reversal is counted.

3. After the initial DFS which counts reversals starting from the root, the dfs2 function is then called in a similar manner. It's called initially with the root node 0 and its parent -1.

   • This function also iterates over each adjacent node j of the current node i and its edge direction k.
   • Instead of just counting reversals, dfs2 propagates the reversal count to each node. It sets ans[j] to ans[i] + k for each node j that is not the parent.
   • The value of k is added because if k is 1, no reversal is needed from i to j, and if k is -1, an extra reversal is needed to reach j from i.

Through this propagation, dfs2 calculates the optimal number of reversals needed to reach all other nodes from a given node i in the tree. Since ans[i] already contains the minimum reversals to reach i from the root (calculated in the initial DFS), adding k adjusts this number based on the direction of the edge leading to the child node j.

In the end, we return the ans array that contains the minimum number of reversals needed for each node to reach all other nodes. The usage of DFS is crucial as it allows us to visit each node and its subtree, accumulating and propagating the count of necessary reversals across the graph. The logic applied in the second DFS makes use of information collected during the first DFS to solve the problem efficiently.

Example Walkthrough

To illustrate the solution approach, consider a small directed graph with n = 3 nodes and the following edges: edges = [[0, 1], [1, 2], [2, 0]]. The graph forms a cycle if we ignore the direction of the edges. Here's how we apply the solution approach to determine the number of reversals needed for each node to reach all other nodes:

1. We initialize the graph representation g with the adjacency list for the directed graph:

   g = [
     [(1, 1)],      // Node 0 has a directed edge towards Node 1
     [(2, 1)],      // Node 1 has a directed edge towards Node 2
     [(0, -1)]      // Node 2 has an edge pointing towards Node 0, which needs reversal
   ]

2. We define a dfs function to traverse the graph and begin by calling dfs(0, -1), which starts the traversal from Node 0 with -1 signifying no parent.

   • The dfs function will find no reversals are needed from Node 0 to Node 1. The traversal continues to Node 2, where the dfs detects that an edge reversal is needed (from Node 2 to Node 0). This increments the number of necessary edge reversals for the root, ans[0], by 1.

3. We then implement the dfs2 function and first call it with dfs2(0, -1) to set the number of reversals for each node. It propagates the reversals needed from Node 0 to Node 1 (which remains 0 since the edge from Node 0 to Node 1 does not require reversal). Then it sets the number of reversals for Node 2:

   • Since no reversal is needed to go from Node 1 to Node 2 (k = 1), ans[2] is set equal to ans[1].
   • However, remember that starting from Node 2, we would need to reverse the edge to Node 0 to create a path from Node 2 to Node 0, adding 1 to the reversal count for Node 2.

By applying these steps, we obtain the final array ans, which gives us ans = [1, 0, 0]. This means that starting from Node 0, one reversal is required for all nodes to be reachable (reversing the edge from Node 2 to Node 0). For Nodes 1 and 2, no reversals are needed because there is a direct path from these nodes to every other node in the graph.

This example shows the basic thinking behind the solution: construct a graph that captures directionality, perform an initial DFS to count reversals from the root, and then propagate that count to all nodes using a second DFS, adjusting the count based on the direction of the incoming edges to each node.

Solution Implementation

Time and Space Complexity

Time Complexity

The given code consists of two Depth First Search (DFS) functions, dfs and dfs2, which are used to traverse the graph represented by the list of edges. Every vertex and every edge is visited exactly once in each DFS operation.

• The first DFS (dfs) traverses the graph to count the minimum number of edge reversals required. It does so by incrementing the answer only if an edge is directed towards the parent (k < 0), which requires a reversal.
• The second DFS (dfs2) computes the final answer array for each node, by updating the answer for each child based on the edge direction (k) with respect to its parent.

The time complexity of each DFS call is O(V + E) where V is the number of vertices (nodes) and E is the number of edges, since it traverses each vertex and edge exactly once. Since we are running two DFS operations sequentially, and each has the same complexity, the total time complexity of the algorithm is O(V + E + V + E) which simplifies to O(V + E) because constants are dropped in Big O notation.

Therefore, the overall time complexity is O(V + E).

Space Complexity

The space complexity of the code involves the storage required for:

• The adjacency list g which has E entries (each edge stored twice, once for each direction, with an additional bit to indicate the original direction).
• The recursion stack for the DFS calls, which in the worst case, if the graph is implemented as a linked list, could go up to O(V).
• The ans array, which is an array of length V.

Combining the above, the space complexity is O(V + 2E) + O(V) + O(V).

Since E can be at most V(V - 1)/2 for a complete graph, the term O(2E) can dominate the space complexity. However, we typically express space complexity in terms of just O(E) or O(V), depending on which is larger. Therefore, the space complexity is generally represented as O(E + V).

In conclusion, the space complexity of the algorithm is O(E + V).

Learn more about how to find time and space complexity quickly using problem constraints.