2923. Find Champion I

Problem Description

In this problem, you're given data representing the results of matches in a tournament. This data is in the form of a 2D boolean matrix grid, where each dimension represents the n teams in the tournament. If grid[i][j] is 1, it means team i is stronger than team j. Conversely, if it's 0, then team i is weaker than team j. The objective is to find out which team is the champion, with the definition of a champion being a team that is stronger than all other teams.

To reiterate, we're looking for a team that has won against every other team. If such a team exists, it will be declared the champion. Importantly, we know that every team will only have a row where they are the winner if they've won every match against others, and we need to identify such a row to find the champion.

Intuition

When thinking about how to solve this problem, one straightforward method that comes to mind is to enumerate each team and check their match outcomes. If we find a team 'i' that has beaten every other team, which in the matrix translates to a row filled with '1's except the diagonal (as a team does not play against itself), then i is the champion.

To implement this solution in an efficient way, we iterate over each team i by going through each row in the matrix. Then, using a comprehension, we check if every element x in the row is 1 for all other teams j, implying that team i has won every match with others.

If such a condition satisfies, then we have found our champion, and the loop can be terminated early by returning the index i. This approach is simple and works because of the rules of the tournament as given in the problem, ensuring that there can be at most one champion.

Solution Approach

The Python solution provided takes advantage of a few Python-specific features. Firstly, it uses list comprehensions - a compact way to process all entries in a list. In the context of the given solution, the comprehension checks if a team has all victories (1) against other teams, which can be checked by looking through the row corresponding to that team in the grid matrix. The condition x == 1 for j, x in enumerate(row) if i != j makes sure we skip the team's own position since a team cannot compete against itself – that's why the check i != j is there.

Here's a breakdown of key elements used in the implementation:

• Enumeration: The enumerate function is used to get both the index (i) and the value (row) from the grid. The index represents the team number, and the row contains the match outcomes for that team.

• List Comprehensions: Comprehensions are used to go through each match result in the row while skipping the match result against themselves, as indicated by i != j.

• All Function: all() is a built-in function in Python that returns True if all elements of the given iterable are true (or if the iterable is empty). In the given context, it is used to check if a team has a row of all wins (1), which would satisfy the condition to be the champion.

• Early Return: Instead of iterating through the entire matrix and then deciding the result, the solution returns the index (the team number) as soon as it finds the row that satisfies the champion criteria. This is efficient because it stops further unnecessary computation once the champion is identified.

To summarize the solution approach as per the problem:

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

This method iterates through each row with index i:

1. For each team i, check through all match outcomes x in the corresponding row.
2. Skip the outcome where a team played against themselves with the condition i != j.
3. If all outcomes x against other teams j are victories (1), then team i is the champion, and we return i.

In this problem, we make some assumptions:

• There can only be one champion.
• A draw is not possible in a match between two teams.
• When a team wins, their corresponding cell is set as 1 and 0 otherwise.

The solution uses a basic iteration of rows in combination with conditional checks and short-circuits as soon as the condition for being a champion is satisfied, making it an efficient solution to find the champion of the tournament.

Example Walkthrough

Let's walk through a small example using a hypothetical 3-team tournament to illustrate the solution approach. Suppose we have the following tournament results in a boolean matrix:

1grid = [
2  [0, 1, 1],  # Team 0's results
3  [0, 0, 1],  # Team 1's results
4  [0, 0, 0]   # Team 2's results
5]

In this matrix:

• grid[0] represents that Team 0 lost to Team 1 (grid[0][1] = 1 indicates Team 1 is stronger than Team 0) and won against Team 2 (grid[0][2] = 1 indicates Team 0 is stronger than Team 2).
• grid[1] represents that Team 1 lost to Team 0 (grid[1][0] = 0 indicates Team 0 is stronger than Team 1) but won against Team 2 (grid[1][2] = 1).
• grid[2] represents that Team 2 lost to both Team 0 (grid[2][0] = 0) and Team 1 (grid[2][1] = 0).

We want to find out which team is the champion, which is the team that won against every other team. This means we are looking for a row in grid which, except for the diagonal entry, contains all 1s.

Let's apply our solution approach:

1. Start by iterating through the grid matrix, getting both the index i and the row.

   • i = 0, row = [0, 1, 1]: We pass over the first entry (since team doesn't compete against itself) and check the rest. Team 0 has one loss (grid[0][0] = 0), so it cannot be the champion.
   • i = 1, row = [0, 0, 1]: Again, we pass over the diagonal entry. Team 1 has a loss, so it cannot be the champion either.
   • i = 2, row = [0, 0, 0]: We ignore the diagonal, but all other entries are 0, so Team 2 has not won any matches.

2. We used the all() Python function to check if all entries in a row, excluding the diagonal, are 1.

Since none of the teams have a row with all 1s, we conclude that there is no champion in this tournament. If there was a team that won against every other team, the all() call would return True for that team's row, and the function would return that team's index as the champion.

Following is what the Python code might look like for this example:

1def findChampion(grid):
2    for i, row in enumerate(grid):
3        if all(x == 1 for j, x in enumerate(row) if i != j):
4            return i
5    return -1  # Using -1 to indicate no champion found
6
7# Testing the example grid
8grid = [
9  [0, 1, 1],
10  [0, 0, 1],
11  [0, 0, 0]
12]
13
14print(findChampion(grid))  # Output will be -1 since no team meets the criteria

As we can see, the function checks each team's winning record against others and concludes correctly based on the problem description. No team satisfies the condition to be the champion in this particular case.

Solution Implementation

Time and Space Complexity

The time complexity of the provided code is O(n^2) where n is the number of teams. This is because for each team i, the code checks whether all other teams j have been defeated by team i. This involves a nested iteration where the inner loop runs n-1 times (because it skips the case where i == j) for each of the n teams, hence the quadratic time complexity.

The space complexity is O(1) as the code only uses a fixed amount of additional memory for variables like i, row, and x. There are no data structures used that grow with the size of the input.

Learn more about how to find time and space complexity quickly using problem constraints.