How do I solve the Find Champion I problem?

To solve Find Champion I, you can use matrix traversal or graph-based algorithms, where you identify the team with no stronger opponents. Common techniques include checking rows and columns or applying a topological sort.

What algorithm is best for Find Champion I?

The best algorithm for Find Champion I typically involves checking each team’s relationships with others in the matrix, using either graph theory or simple matrix traversal.

What is the time complexity of Find Champion I?

The time complexity of the problem is O(n^2), where n is the number of teams, since you're typically scanning a 2D matrix of size n x n.

Can the Find Champion I problem be solved with a graph?

Yes, Find Champion I can be solved by representing the tournament as a directed graph and finding the node (team) with no incoming edges.

What are some common mistakes in Find Champion I?

Common mistakes include not accounting for the asymmetry of the matrix or overcomplicating the solution with unnecessary data structures or inefficient algorithms.

#2923

Easy

auto_awesome数组·matrix

LeetCode 题解工作台

找到冠军 I

一场比赛中共有 n 支队伍，按从 0 到 n - 1 编号。给你一个下标从 0 开始、大小为 n * n 的二维布尔矩阵 grid 。对于满足 0 且 i != j 的所有 i, j ：如果 grid[i][j] == 1 ，那么 i 队比 j 队强；否则， j 队比 i 队强。在这场比…

数组矩阵

题目描述

一场比赛中共有 n 支队伍，按从 0 到 n - 1 编号。

给你一个下标从 0 开始、大小为 n * n 的二维布尔矩阵 grid 。对于满足 0 <= i, j <= n - 1 且 i != j 的所有 i, j ：如果 grid[i][j] == 1，那么 i 队比 j 队强；否则，j 队比 i 队强。

在这场比赛中，如果不存在某支强于 a 队的队伍，则认为 a 队将会是冠军。

返回这场比赛中将会成为冠军的队伍。

示例 1：

输入：grid = [[0,1],[0,0]]
输出：0
解释：比赛中有两支队伍。
grid[0][1] == 1 表示 0 队比 1 队强。所以 0 队是冠军。

示例 2：

输入：grid = [[0,0,1],[1,0,1],[0,0,0]]
输出：1
解释：比赛中有三支队伍。
grid[1][0] == 1 表示 1 队比 0 队强。
grid[1][2] == 1 表示 1 队比 2 队强。
所以 1 队是冠军。

提示：

n == grid.length
n == grid[i].length
2 <= n <= 100
grid[i][j] 的值为 0 或 1
对于所有 i， grid[i][i] 等于 0.
对于满足 i != j 的所有 i, j ，grid[i][j] != grid[j][i] 均成立
生成的输入满足：如果 a 队比 b 队强，b 队比 c 队强，那么 a 队比 c 队强

lightbulb

解题思路

方法一：枚举

我们可以枚举每一支队伍 $i$ ，如果 $i$ 队的每一场比赛都赢了，那么 $i$ 队就是冠军，直接返回 $i$ 即可。

时间复杂度 $O(n^2)$ ，其中 $n$ 是队伍数量。空间复杂度 $O(1)$ 。

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class Solution:
    def findChampion(self, grid: List[List[int]]) -> int:
        for i, row in enumerate(grid):
            if all(x == 1 for j, x in enumerate(row) if i != j):
                return i

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Look for candidates who can efficiently implement a solution using graph traversal techniques or matrix analysis.
question_mark
Candidates should understand the problem's matrix-based structure and avoid unnecessary brute-force approaches.
question_mark
Assess if the candidate can choose the most efficient algorithm depending on the given matrix size and constraints.

warning

常见陷阱

外企场景

error
Failing to recognize the asymmetry of the matrix and attempting to solve it with incorrect assumptions about the data structure.
error
Overcomplicating the solution by introducing unnecessary data structures or algorithms.
error
Not optimizing the solution for larger values of n, leading to inefficient runtime for the problem's constraints.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Use a directed graph representation and explore different traversal methods to identify the champion.
arrow_right_alt
Change the matrix to include additional constraints, such as teams with partial strengths against each other.
arrow_right_alt
Introduce a scenario where ties can occur, and the algorithm must handle determining the champion under these conditions.

help

常见问题

外企场景

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