#2923

Easy

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Find Champion I

Find Champion I is an easy problem focusing on identifying the strongest team in a tournament using matrix-based comparisons.

array matrix

Problem Statement

In a tournament with n teams, each team is represented by a number from 0 to n - 1. A 2D boolean matrix grid of size n * n is provided, where grid[i][j] = 1 means team i is stronger than team j, and grid[i][j] = 0 means team j is stronger than team i. The matrix is asymmetrical, meaning grid[i][j] != grid[j][i] for all i != j.

A team is considered the champion if no other team is stronger than it, meaning there is no team b such that grid[b][a] == 1. Your task is to find the champion team from this matrix.

Examples

Example 1

Input: grid = [[0,1],[0,0]]

Output: 0

There are two teams in this tournament. grid[0][1] == 1 means that team 0 is stronger than team 1. So team 0 will be the champion.

Example 2

Input: grid = [[0,0,1],[1,0,1],[0,0,0]]

Output: 1

There are three teams in this tournament. grid[1][0] == 1 means that team 1 is stronger than team 0. grid[1][2] == 1 means that team 1 is stronger than team 2. So team 1 will be the champion.

Constraints

n == grid.length
n == grid[i].length
2 <= n <= 100
grid[i][j] is either 0 or 1.
For all i grid[i][i] is 0.
For all i, j that i != j, grid[i][j] != grid[j][i].
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 Approach

Graph Theory Approach

Treat the matrix as a directed graph where each team is a node, and a directed edge from team i to team j indicates that team i is stronger than team j. The champion will be the node with no incoming edges. This method can be efficient for solving the problem by traversing the matrix and counting the incoming edges for each team.

Row and Column Check

A straightforward method involves checking each row and column in the matrix. If a team i has no '0's in its row and no '1's in its column, it is the champion. This approach focuses on verifying that a team is stronger than all others without being beaten.

Topological Sort

Another approach is to apply a topological sort to the matrix's graph representation. If we can topologically sort the teams such that one team comes out as the last in the order, that team will be the champion, as no team in the tournament can beat it.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the approach chosen. A matrix scan for a row and column check is O(n^2), while topological sorting can be O(n^2) in the worst case. The space complexity for both methods is O(n), since we're only storing the matrix or an auxiliary data structure like an adjacency list for graph traversal.

What Interviewers Usually Probe

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

Common Pitfalls or Variants

Common pitfalls

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

Follow-up variants

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

FAQ

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.

terminal

Solution

Solution 1: Enumeration

We can enumerate each team $i$. If team $i$ has won every match, then team $i$ is the champion, and we can directly return $i$.

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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

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