How do you solve Find if Path Exists in Graph on LeetCode?

Find if Path Exists in Graph (LeetCode #1971) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding graph traversal techniques is crucial.

What is the brute force solution for Find if Path Exists in Graph?

The brute force approach for Find if Path Exists in Graph is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible paths from the source to the destination. This can be done using a depth-first search (DFS) or breadth-first search (BFS) to explore all connections from the source vertex. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find if Path Exists in Graph?

Find if Path Exists in Graph uses the following concepts: Depth-First Search, Breadth-First Search, Union-Find, Graph Theory. The recommended patterns to study are: Graph Traversal, Breadth-First Search, Depth-First Search.

What are the interview tips for Find if Path Exists in Graph?

Explain your thought process clearly while coding. Consider edge cases, such as disconnected graphs. Practice both DFS and BFS approaches.

What are common mistakes when solving Find if Path Exists in Graph?

Not handling visited nodes correctly, leading to infinite loops. Confusing directed and undirected graphs.

#1971

Find if Path Exists in Graph

Easy

Depth-First SearchBreadth-First SearchUnion-FindGraph TheoryGraph TraversalBreadth-First SearchDepth-First Search

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses BFS or DFS to traverse the graph efficiently, ensuring that we only explore each vertex once. This reduces unnecessary checks and improves performance significantly.

⚙️

Algorithm

3 steps

1Step 1: Build an adjacency list from the edges to represent the graph.
2Step 2: Initialize a queue (for BFS) or a stack (for DFS) and a visited set.
3Step 3: Start from the source, add it to the queue/stack, and explore all reachable vertices until the destination is found or all options are exhausted.

solution.py20 lines

1# Full working Python code
2from collections import defaultdict, deque
3
4def validPath(n, edges, source, destination):
5    graph = defaultdict(list)
6    for u, v in edges:
7        graph[u].append(v)
8        graph[v].append(u)
9
10    visited = set()
11    queue = deque([source])
12    while queue:
13        node = queue.popleft()
14        if node == destination:
15            return True
16        visited.add(node)
17        for neighbor in graph[node]:
18            if neighbor not in visited:
19                queue.append(neighbor)
20    return False

ℹ

Complexity note: The time complexity is O(n) because we traverse each vertex and edge once. The space complexity is O(n) due to the storage of the graph in an adjacency list and the visited set.

1Understanding graph traversal techniques is crucial.
2Building an adjacency list is a common pattern in graph problems.

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