Introduction

In this blog post, we will explore how to solve the LeetCode problem "2924" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

Here's the solution to the problem:

• High-level approach:

  • Calculate the in-degree of each node (team). The in-degree of a node represents the number of teams that are stronger than it.
  • A champion is a node with an in-degree of 0, meaning no other team is stronger.
  • If there's exactly one node with an in-degree of 0, that's the champion. Otherwise, there's no unique champion.

• Complexity:

  • Runtime: O(m + n), where m is the number of edges and n is the number of teams.
  • Storage: O(n)

Code

    def findChampion(n: int, edges: list[list[int]]) -> int:
    """
    Finds the champion of the tournament if there is a unique champion, otherwise, return -1.
    """
    in_degree = [0] * n
    for edge in edges:
        in_degree[edge[1]] += 1

    champions = []
    for i in range(n):
        if in_degree[i] == 0:
            champions.append(i)

    if len(champions) == 1:
        return champions[0]
    else:
        return -1