Leetcode #2924: Find Champion II
In this guide, we solve Leetcode #2924 Find Champion 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
There are n teams numbered from 0 to n - 1 in a tournament; each team is also a node in a DAG. You are given the integer n and a 0-indexed 2D integer array edges of length m representing the DAG, where edges[i] = [ui, vi] indicates that there is a directed edge from team ui to team vi in the graph.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: 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]]
Output: 0
Explanation: Team 1 is weaker than team 0. Team 2 is weaker than team 1. So the champion is team 0.
Python Solution
class Solution:
def findChampion(self, n: int, edges: List[List[int]]) -> int:
indeg = [0] * n
for _, v in edges:
indeg[v] += 1
return -1 if indeg.count(0) != 1 else indeg.index(0)
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.