2024-11-26 Daily Challenge

Today I have done leetcode's November LeetCoding Challenge with cpp.

November LeetCoding Challenge 26

Description

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.

Solution

class Solution {
public:
  int findChampion(int n, vector<vector<int>>& edges) {
    vector<int> degree(n);

    for(const auto &edge : edges) {
      degree[edge[1]] += 1;
    }
    int answer = -1;
    for(int i = 0; i < n; ++i) {
      if(!degree[i]) {
        if(answer == -1) answer = i;
        else return -1;
      }
    }
    return answer;
  }
};