2924 - 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.

C++

• class Solution {
  public:
      int findChampion(int n, vector<vector<int>>& edges) {
          int indeg[n];
          memset(indeg, 0, sizeof(indeg));
          for (auto& e : edges) {
              ++indeg[e[1]];
          }
          int ans = -1, cnt = 0;
          for (int i = 0; i < n; ++i) {
              if (indeg[i] == 0) {
                  ++cnt;
                  ans = i;
              }
          }
          return cnt == 1 ? ans : -1;
      }
  };

JAVA

• class Solution {
      public int findChampion(int n, int[][] edges) {
          int[] indeg = new int[n];
          for (var e : edges) {
              ++indeg[e[1]];
          }
          int ans = -1, cnt = 0;
          for (int i = 0; i < n; ++i) {
              if (indeg[i] == 0) {
                  ++cnt;
                  ans = i;
              }
          }
          return cnt == 1 ? ans : -1;
      }
  }

PYTHON

• class Solution:
      def findChampion(self, n, edges):
          indeg = [0] * n
          for e in edges:
              indeg[e[1]] += 1
  
          ans, cnt = -1, 0
          for i in range(n):
              if indeg[i] == 0:
                  cnt += 1
                  ans = i
  
          return ans if cnt == 1 else -1