Leetcode #1786: Number of Restricted Paths From First to Last Node
In this guide, we solve Leetcode #1786 Number of Restricted Paths From First to Last Node in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
There is an undirected weighted connected graph. You are given a positive integer n which denotes that the graph has n nodes labeled from 1 to n, and an array edges where each edges[i] = [ui, vi, weighti] denotes that there is an edge between nodes ui and vi with weight equal to weighti.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Graph, Topological Sort, Dynamic Programming, Shortest Path, Heap (Priority Queue)
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: n = 5, edges = [[1,2,3],[1,3,3],[2,3,1],[1,4,2],[5,2,2],[3,5,1],[5,4,10]]
Output: 3
Explanation: Each circle contains the node number in black and its distanceToLastNode value in blue. The three restricted paths are:
1) 1 --> 2 --> 5
2) 1 --> 2 --> 3 --> 5
3) 1 --> 3 --> 5
Python Solution
class Solution:
def countRestrictedPaths(self, n: int, edges: List[List[int]]) -> int:
def dfs(i):
if i == n:
return 1
ans = 0
for j, _ in g[i]:
if dist[i] > dist[j]:
ans = (ans + dfs(j)) % mod
return ans
g = defaultdict(list)
for u, v, w in edges:
g[u].append((v, w))
g[v].append((u, w))
q = [(0, n)]
dist = [inf] * (n + 1)
dist[n] = 0
mod = 10**9 + 7
while q:
_, u = heappop(q)
for v, w in g[u]:
if dist[v] > dist[u] + w:
dist[v] = dist[u] + w
heappush(q, (dist[v], v))
return dfs(1)
Complexity
The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.