Leetcode #332: Reconstruct Itinerary
In this guide, we solve Leetcode #332 Reconstruct Itinerary 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
You are given a list of airline tickets where tickets[i] = [fromi, toi] represent the departure and the arrival airports of one flight. Reconstruct the itinerary in order and return it.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Depth-First Search, Graph, Eulerian Circuit
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: tickets = [["MUC","LHR"],["JFK","MUC"],["SFO","SJC"],["LHR","SFO"]]
Output: ["JFK","MUC","LHR","SFO","SJC"]
Python Solution
class Solution:
def findItinerary(self, tickets: List[List[str]]) -> List[str]:
def dfs(f: str):
while g[f]:
dfs(g[f].pop())
ans.append(f)
g = defaultdict(list)
for f, t in sorted(tickets, reverse=True):
g[f].append(t)
ans = []
dfs("JFK")
return ans[::-1]
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.