Question

2924. Find Champion II

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Description

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.

Solutions

Solution 1: Counting In-degrees

Based on the problem description, we only need to count the in-degrees of each node and record them in an array $indeg$. If only one node has an in-degree of $0$, then this node is the champion; otherwise, there is no unique champion.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes.

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)

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;
  }
}

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;
  }
};

func findChampion(n int, edges [][]int) int {
  indeg := make([]int, n)
  for _, e := range edges {
    indeg[e[1]]++
  }
  ans, cnt := -1, 0
  for i, x := range indeg {
    if x == 0 {
      cnt++
      ans = i
    }
  }
  if cnt == 1 {
    return ans
  }
  return -1
}

function findChampion(n: number, edges: number[][]): number {
  const indeg: number[] = Array(n).fill(0);
  for (const [_, v] of edges) {
    ++indeg[v];
  }
  let [ans, cnt] = [-1, 0];
  for (let i = 0; i < n; ++i) {
    if (indeg[i] === 0) {
      ++cnt;
      ans = i;
    }
  }
  return cnt === 1 ? ans : -1;
}