Leetcode #1724: Checking Existence of Edge Length Limited Paths II
In this guide, we solve Leetcode #1724 Checking Existence of Edge Length Limited Paths II 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
An undirected graph of n nodes is defined by edgeList, where edgeList[i] = [ui, vi, disi] denotes an edge between nodes ui and vi with distance disi. Note that there may be multiple edges between two nodes, and the graph may not be connected.
Quick Facts
- Difficulty: Hard
- Premium: Yes
- Tags: Union Find, Graph, Minimum Spanning Tree
Intuition
The data forms a graph, so we should explore nodes and edges systematically.
A traversal ensures we visit each node once while maintaining the needed state.
Approach
Build an adjacency list and traverse with BFS or DFS.
Aggregate results as you visit nodes.
Steps:
- Build the graph.
- Traverse with BFS/DFS.
- Accumulate the required output.
Example
Input
["DistanceLimitedPathsExist", "query", "query", "query", "query"]
[[6, [[0, 2, 4], [0, 3, 2], [1, 2, 3], [2, 3, 1], [4, 5, 5]]], [2, 3, 2], [1, 3, 3], [2, 0, 3], [0, 5, 6]]
Output
[null, true, false, true, false]
Explanation
DistanceLimitedPathsExist distanceLimitedPathsExist = new DistanceLimitedPathsExist(6, [[0, 2, 4], [0, 3, 2], [1, 2, 3], [2, 3, 1], [4, 5, 5]]);
distanceLimitedPathsExist.query(2, 3, 2); // return true. There is an edge from 2 to 3 of distance 1, which is less than 2.
distanceLimitedPathsExist.query(1, 3, 3); // return false. There is no way to go from 1 to 3 with distances strictly less than 3.
distanceLimitedPathsExist.query(2, 0, 3); // return true. There is a way to go from 2 to 0 with distance < 3: travel from 2 to 3 to 0.
distanceLimitedPathsExist.query(0, 5, 6); // return false. There are no paths from 0 to 5.
Python Solution
class PersistentUnionFind:
def __init__(self, n):
self.rank = [0] * n
self.p = list(range(n))
self.version = [inf] * n
def find(self, x, t=inf):
if self.p[x] == x or self.version[x] >= t:
return x
return self.find(self.p[x], t)
def union(self, a, b, t):
pa, pb = self.find(a), self.find(b)
if pa == pb:
return False
if self.rank[pa] > self.rank[pb]:
self.version[pb] = t
self.p[pb] = pa
else:
self.version[pa] = t
self.p[pa] = pb
if self.rank[pa] == self.rank[pb]:
self.rank[pb] += 1
return True
class DistanceLimitedPathsExist:
def __init__(self, n: int, edgeList: List[List[int]]):
self.puf = PersistentUnionFind(n)
edgeList.sort(key=lambda x: x[2])
for u, v, dis in edgeList:
self.puf.union(u, v, dis)
def query(self, p: int, q: int, limit: int) -> bool:
return self.puf.find(p, limit) == self.puf.find(q, limit)
Complexity
The time complexity is O(V+E). The space complexity is O(V).
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.