Leetcode #1971: Find if Path Exists in Graph
In this guide, we solve Leetcode #1971 Find if Path Exists in Graph 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 a bi-directional graph with n vertices, where each vertex is labeled from 0 to n - 1 (inclusive). The edges in the graph are represented as a 2D integer array edges, where each edges[i] = [ui, vi] denotes a bi-directional edge between vertex ui and vertex vi.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Depth-First Search, Breadth-First Search, Union Find, Graph
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: n = 3, edges = [[0,1],[1,2],[2,0]], source = 0, destination = 2
Output: true
Explanation: There are two paths from vertex 0 to vertex 2:
- 0 → 1 → 2
- 0 → 2
Python Solution
class Solution:
def validPath(
self, n: int, edges: List[List[int]], source: int, destination: int
) -> bool:
def dfs(i: int) -> bool:
if i == destination:
return True
vis.add(i)
for j in g[i]:
if j not in vis and dfs(j):
return True
return False
g = [[] for _ in range(n)]
for u, v in edges:
g[u].append(v)
g[v].append(u)
vis = set()
return dfs(source)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
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.