3189 Find Champion Ii

Find Champion II

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.

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

        for k in edges:
            teams[k[1]] = 1
        #print(teams)
        result = -1
        for i, k in enumerate(teams):
            if k == 0 and result == -1:
                result = i
            
            elif k == 0 and result != -1:
                return -1

        return result