How do you solve Number of Restricted Paths From First to Last Node on LeetCode?

Number of Restricted Paths From First to Last Node (LeetCode #1786) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O((n + e) log n) time complexity and O(n + e) space complexity. Key insight: Understanding Dijkstra's algorithm is crucial for finding shortest paths in graphs.

What is the time complexity of Number of Restricted Paths From First to Last Node?

The optimal solution for Number of Restricted Paths From First to Last Node runs in O((n + e) log n) time and uses O(n + e) space. The time complexity is O((n + e) log n) due to Dijkstra's algorithm, where e is the number of edges. The space complexity is O(n + e) for storing the graph and distances.

What is the brute force solution for Number of Restricted Paths From First to Last Node?

The brute force approach for Number of Restricted Paths From First to Last Node is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach explores all possible paths from node 1 to node n, checking each path to see if it is restricted. This method, while straightforward, can be very slow for larger graphs due to its exhaustive nature. The optimal Optimal Solution approach improves this to O((n + e) log n).

What are the interview tips for Number of Restricted Paths From First to Last Node?

Explain your thought process clearly while coding. Discuss edge cases and how your solution handles them. Practice explaining your code to someone else to reinforce your understanding.

What are common mistakes when solving Number of Restricted Paths From First to Last Node?

Failing to correctly implement Dijkstra's algorithm. Not considering edge cases where there are no paths.

#1786

Number of Restricted Paths From First to Last Node

Medium

Dynamic ProgrammingGraph TheoryTopological SortHeap (Priority Queue)Shortest PathGraph TraversalDynamic ProgrammingDijkstra's Algorithm

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O((n + e) log n)
Space	O(1)	O(n + e)

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Intuition

Time O((n + e) log n)Space O(n + e)

The optimal solution uses Dijkstra's algorithm to find the shortest distances from the last node to all other nodes, then counts valid paths using a dynamic programming approach. This is efficient because it avoids redundant calculations and leverages the properties of shortest paths.

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Algorithm

4 steps

1Step 1: Use Dijkstra's algorithm to calculate the shortest distance from node n to all other nodes.
2Step 2: Create a dynamic programming array where dp[i] represents the number of restricted paths from node i to node n.
3Step 3: Traverse the nodes in reverse order (from n-1 to 1) and for each node, check its neighbors. If the distance to a neighbor is less than the current node, update dp for that neighbor.
4Step 4: Return dp[1] as the result.

solution.py35 lines

1# Full working Python code
2import heapq
3from collections import defaultdict
4
5def countRestrictedPaths(n, edges):
6    graph = defaultdict(list)
7    for u, v, w in edges:
8        graph[u].append((v, w))
9        graph[v].append((u, w))
10
11    distances = dijkstra(n, graph)
12    dp = [0] * (n + 1)
13    dp[n] = 1
14
15    for node in range(n - 1, 0, -1):
16        for neighbor, weight in graph[node]:
17            if distances[node] > distances[neighbor]:
18                dp[node] += dp[neighbor]
19                dp[node] %= 10**9 + 7
20
21    return dp[1]
22
23def dijkstra(n, graph):
24    distances = [float('inf')] * (n + 1)
25    distances[n] = 0
26    min_heap = [(0, n)]
27
28    while min_heap:
29        curr_distance, node = heapq.heappop(min_heap)
30        for neighbor, weight in graph[node]:
31            if curr_distance + weight < distances[neighbor]:
32                distances[neighbor] = curr_distance + weight
33                heapq.heappush(min_heap, (distances[neighbor], neighbor))
34
35    return distances

ℹ

Complexity note: The time complexity is O((n + e) log n) due to Dijkstra's algorithm, where e is the number of edges. The space complexity is O(n + e) for storing the graph and distances.

1Understanding Dijkstra's algorithm is crucial for finding shortest paths in graphs.
2Dynamic programming can significantly optimize counting problems by avoiding redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.