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 == 30 <= ui < vi < n0 <= cnti <= 1040 <= maxMoves <= 1091 <= n <= 3000When 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:
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:
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_countThis 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:
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| Case | How to Handle |
|---|---|
| Empty edges list | Return 0 since no edges exist, thus no reachable nodes beyond the initial node. |
| edges list with a single edge and maxMoves is 0 | Return 1, because only the starting node (0) is reachable. |
| edges list with high subdivision count and large maxMoves | 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) | The algorithm should function correctly, essentially exploring the original graph if subdivisions are zero. |
| Edges forming a cycle in the original graph before subdivision | Dijkstra's algorithm handles cycles correctly as it always chooses the shortest path. |
| maxMoves exceeding the practical limit due to integer overflow during calculations | 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. | 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. | Only reachable nodes from node 0 are counted, so disconnected components are automatically ignored after Dijkstra's. |