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Reachable Nodes In Subdivided Graph

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Graphs

You are given an undirected graph (the "original graph") with n nodes labeled from 0 to n - 1. You decide to subdivide each edge in the graph into a chain of nodes, with the number of new nodes varying between each edge.

The graph is given as a 2D array of edges where edges[i] = [ui, vi, cnti] indicates that there is an edge between nodes ui and vi in the original graph, and cnti is the total number of new nodes that you will subdivide the edge into. Note that cnti == 0 means you will not subdivide the edge.

To subdivide the edge [ui, vi], replace it with (cnti + 1) new edges and cnti new nodes. The new nodes are x1, x2, ..., xcnti, and the new edges are [ui, x1], [x1, x2], [x2, x3], ..., [xcnti-1, xcnti], [xcnti, vi].

In this new graph, you want to know how many nodes are reachable from the node 0, where a node is reachable if the distance is maxMoves or less.

Given the original graph and maxMoves, return the number of nodes that are reachable from node 0 in the new graph.

Example 1:

Input: edges = [[0,1,10],[0,2,1],[1,2,2]], maxMoves = 6, n = 3
Output: 13
Explanation: The edge subdivisions are shown in the image above.
The nodes that are reachable are highlighted in yellow.

Example 2:

Input: edges = [[0,1,4],[1,2,6],[0,2,8],[1,3,1]], maxMoves = 10, n = 4
Output: 23

Example 3:

Input: edges = [[1,2,4],[1,4,5],[1,3,1],[2,3,4],[3,4,5]], maxMoves = 17, n = 5
Output: 1
Explanation: Node 0 is disconnected from the rest of the graph, so only node 0 is reachable.

Constraints:

  • 0 <= edges.length <= min(n * (n - 1) / 2, 104)
  • edges[i].length == 3
  • 0 <= ui < vi < n
  • There are no multiple edges in the graph.
  • 0 <= cnti <= 104
  • 0 <= maxMoves <= 109
  • 1 <= n <= 3000

Solution


Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

  1. What are the ranges for the values of `n`, `edges.length`, `u`, `v`, and `cnt`? Should I be concerned about integer overflow?
  2. Are the edge subdivisions permanent, or only for the purpose of counting reachable nodes within the given `maxMoves`?
  3. Is the input graph guaranteed to be connected after the subdivisions, or could there be unreachable sections?
  4. Are there any restrictions on the structure of the graph? For instance, is it guaranteed to be acyclic after subdivision?
  5. If there are multiple paths to a node and `maxMoves` is sufficient to reach it via multiple paths, do I count it only once?

Brute Force Solution

Approach

To find how many nodes we can reach, imagine exploring the graph step by step. The brute force method involves simulating every possible path we can take until we run out of moves, then simply counting all the different nodes we've managed to reach.

Here's how the algorithm would work step-by-step:

  1. Start at the initial node (node zero).
  2. Consider all the possible edges that lead out from the current node.
  3. For each edge, determine the number of new nodes that are added when following that edge.
  4. If we have enough moves left, 'walk' along that edge, updating the number of moves remaining and adding all the nodes on that edge to our set of reached nodes.
  5. Repeat the process from each newly reached node, considering all possible edges from those nodes.
  6. Continue exploring until we run out of moves or there are no more new nodes to reach.
  7. Finally, count all the unique nodes we've encountered during this exploration. This gives us the total number of reachable nodes.

Code Implementation

def reachable_nodes_brute_force(edges, max_moves, number_of_nodes):
    reachable_nodes = {0}
    queue = [(0, max_moves)]

    while queue:
        current_node, remaining_moves = queue.pop(0)

        for start_node, end_node, number_of_new_nodes in edges:
            if start_node == current_node:
                # Check if we have enough moves to traverse the edge.
                if remaining_moves >= number_of_new_nodes + 1:

                    new_remaining_moves = remaining_moves - (number_of_new_nodes + 1)
                    for i in range(1, number_of_new_nodes + 1):
                        intermediate_node = (start_node, end_node, i)
                        reachable_nodes.add(intermediate_node)

                    if end_node not in reachable_nodes:
                        reachable_nodes.add(end_node)
                        # Only add to the queue if the end node hasn't
                        # been reached with more moves remaining.
                        queue.append((end_node, new_remaining_moves))

            elif end_node == current_node:
                # Check if we have enough moves to traverse the edge.
                if remaining_moves >= number_of_new_nodes + 1:

                    new_remaining_moves = remaining_moves - (number_of_new_nodes + 1)
                    for i in range(1, number_of_new_nodes + 1):
                        intermediate_node = (start_node, end_node, i)
                        reachable_nodes.add(intermediate_node)

                    if start_node not in reachable_nodes:
                        reachable_nodes.add(start_node)
                        # Only add to the queue if the start node hasn't
                        # been reached with more moves remaining.
                        queue.append((start_node, new_remaining_moves))

    # Sum of normal nodes that can be reached.
    normal_node_count = sum(1 for node in reachable_nodes if isinstance(node, int))
    # Sum of subdivided nodes that can be reached.
    subdivided_node_count = sum(1 for node in reachable_nodes if not isinstance(node, int))

    return normal_node_count + subdivided_node_count

Big(O) Analysis

Time Complexity
O(M * E)The algorithm explores the graph by iterating through possible moves. In the worst case, we might explore each edge multiple times, potentially up to our maximum moves limit M. For each move, we iterate through all the edges connected to the current node. Therefore, if E is the total number of edges in the subdivided graph, the algorithm could potentially visit or consider M * E edge traversals during the exploration process. Consequently, the time complexity is approximately O(M * E), where M is the maximum moves and E is the number of edges.
Space Complexity
O(E + V)The brute force approach, as described, essentially performs a graph traversal (likely a form of Depth First Search or Breadth First Search, although not explicitly stated). To keep track of visited nodes and avoid cycles, we use a set to store reached nodes, which can grow up to the size of all possible nodes in the graph (V). Additionally, the 'walking' along edges and 'considering all possible edges' implicitly uses a queue or a recursion stack (depending on if it is BFS or DFS) to manage nodes to explore, which, in the worst-case scenario can store all the edges (E) currently being considered. Therefore, the auxiliary space used is proportional to the sum of the number of vertices (V) and edges (E) in the graph, resulting in a space complexity of O(E + V).

Optimal Solution

Approach

This problem asks us to find how many nodes are reachable in a graph after some edges have been split into smaller segments. The trick is to focus on exploring the graph efficiently and avoid recounting the nodes we have already reached by prioritizing exploration near the starting point.

Here's how the algorithm would work step-by-step:

  1. Imagine you're exploring a network of paths, starting from a central location.
  2. Keep track of how much 'energy' you have to spend on walking these paths, where each segment you walk uses up some energy.
  3. Prioritize exploring the closest paths first, so you use your limited energy wisely and reach as many nodes as possible near your starting point.
  4. As you explore, remember which nodes you've already visited to avoid wasting energy going back over the same ground.
  5. When you encounter a path that has been divided into segments, count each new segment you explore along that path.
  6. Keep going until you run out of energy or have explored all reachable nodes within your energy limit.
  7. The total count of nodes you've visited, including the original nodes and the nodes created by splitting the paths, is your answer.

Code Implementation

import heapq

def reachableNodes(edges, max_moves, number_of_nodes):
    adjacency_list = [[] for _ in range(number_of_nodes)]
    for u, v, weight in edges:
        adjacency_list[u].append((v, weight + 1))
        adjacency_list[v].append((u, weight + 1))

    # Priority queue to store nodes to visit next, by distance from start
    priority_queue = [(0, 0)]
    distances = {0: 0}
    reachable_nodes = 0

    while priority_queue:
        distance, current_node = heapq.heappop(priority_queue)

        if distance > distances.get(current_node, float('inf')):
            continue

        reachable_nodes += 1

        # Explore neighbors of the current node
        for neighbor, edge_weight in adjacency_list[current_node]:
            new_distance = distance + edge_weight

            if new_distance <= max_moves:
                if new_distance < distances.get(neighbor, float('inf')):
                    distances[neighbor] = new_distance
                    heapq.heappush(priority_queue, (new_distance, neighbor))

    subdivided_nodes_reached = 0
    for u, v, edge_weight in edges:
        # Calculate how many nodes in the edge (u, v) can be reached
        distance_u = distances.get(u, float('inf'))
        distance_v = distances.get(v, float('inf'))

        reachable_from_u = max(0, max_moves - distance_u)
        reachable_from_v = max(0, max_moves - distance_v)

        # Avoid double-counting nodes
        subdivided_nodes_reached += min(edge_weight, reachable_from_u + reachable_from_v)

    # We want to count all nodes reachable given maxMoves.
    return reachable_nodes + subdivided_nodes_reached

Big(O) Analysis

Time Complexity
O(E log(V) + R)The algorithm uses a priority queue to explore the graph, where E is the number of edges and V is the number of vertices (original nodes and subdivided nodes). Each edge can be visited and potentially added to the priority queue. The priority queue operations (insert and extract min) take O(log(V)) time, resulting in O(E log(V)) complexity. Additionally, we iterate through the edges of the graph during the initial setup. Finally, R represents the number of reachable nodes, which contributes linearly to the overall time complexity as we count each reachable node. Therefore, the overall time complexity is dominated by O(E log(V) + R).
Space Complexity
O(E + N)The algorithm uses a priority queue to explore nodes, which in the worst case can hold all edges resulting from subdivisions, contributing O(E) space where E is the number of edges after subdivision. We also need a set or similar data structure to keep track of visited nodes to avoid revisiting them, which can grow up to the number of nodes in the graph including the newly created ones from subdivision, contributing O(N) space where N is the total number of nodes after subdivisions. Therefore, the total auxiliary space complexity is O(E + N). N could potentially be larger than the original number of nodes given the subdivision, contributing to overall memory usage.

Edge Cases

Empty edges list
How to Handle:
Return 0 since no edges exist, thus no reachable nodes beyond the initial node.
edges list with a single edge and maxMoves is 0
How to Handle:
Return 1, because only the starting node (0) is reachable.
edges list with high subdivision count and large maxMoves
How to Handle:
Use Dijkstra's or a similar algorithm with a priority queue to efficiently find the shortest paths within the allowed moves.
edges with zero subdivisions (original graph)
How to Handle:
The algorithm should function correctly, essentially exploring the original graph if subdivisions are zero.
Edges forming a cycle in the original graph before subdivision
How to Handle:
Dijkstra's algorithm handles cycles correctly as it always chooses the shortest path.
maxMoves exceeding the practical limit due to integer overflow during calculations
How to Handle:
Use appropriate data types (e.g., long) to prevent integer overflows when calculating distances and number of reachable nodes.
All subdivisions are max value and maxMoves is also max value.
How to Handle:
Efficiently process the path lengths in Dijkstra's algorithm or similar, ensuring no integer overflow or excessive memory usage occurs.
Disconnected components exist in the original graph.
How to Handle:
Only reachable nodes from node 0 are counted, so disconnected components are automatically ignored after Dijkstra's.