LeetCode 2924: Find Champion II

Problem Description

LeetCode Problem 2924

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.

A directed edge from a to b in the graph means that team a is stronger than team b and team b is weaker than team a.

Team a will be the champion of the tournament if there is no team b that is stronger than team a.

Return the team that will be the champion of the tournament if there is a unique champion, otherwise, return -1.

Notes

• A cycle is a series of nodes a1, a2, …, an, an+1 such that node a1 is the same node as node an+1, the nodes a1, a2, …, an are distinct, and there is a directed edge from the node ai to node ai+1 for every i in the range [1, n].
• A DAG is a directed graph that does not have any cycle.

Example 1:

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.

Example 2:

Input: n = 4, edges = [[0,2],[1,3],[1,2]]
Output: -1
Explanation: Team 2 is weaker than team 0 and team 1. Team 3 is weaker than team 1. But team 1 and team 0 are not weaker than any other teams. So the answer is -1.

Constraints:

• 1 <= n <= 100
• m == edges.length
• 0 <= m <= n * (n - 1) / 2
• edges[i].length == 2
• 0 <= edge[i][j] <= n - 1
• edges[i][0] != edges[i][1]
• The input is generated such that if team a is stronger than team b, team b is not stronger than team a.
• The input is generated such that if team a is stronger than team b and team b is stronger than team c, then team a is stronger than team c.

Difficulty: Medium

Tags: graph

Rating: 91.67%

Complexity Analysis

The solution has the following complexity characteristics:

• Time Complexity: $O(n + m)$ where $n$ is the number of teams and $m$ is the number of edges
• Space Complexity: $O(n)$ for the incoming array

Solution

Here’s my Python solution to this problem:

class Solution:
    def findChampion(self, n: int, edges: List[List[int]]) -> int:
        incoming = [0] * n

        for start, end in edges:
            incoming[end] += 1

        o = []
        for i, count in enumerate(incoming):
            if not count:
                o.append(i)
        
        if len(o) > 1: return -1
        return o[0]