How do you solve Find Champion I on LeetCode?

Find Champion I (LeetCode #2923) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The champion must be stronger than all other teams, which allows us to eliminate teams quickly.

What is the time complexity of Find Champion I?

The optimal solution for Find Champion I runs in O(n) time and uses O(1) space. This complexity is optimal because we only make a single pass through the teams, resulting in linear time complexity.

What is the brute force solution for Find Champion I?

The brute force approach for Find Champion I is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we check each team to see if there is any other team that is stronger than it. If we find a team that is stronger, we discard the current team as a potential champion. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Champion I?

Find Champion I uses the following concepts: Array, Matrix. The recommended patterns to study are: Array, Matrix.

What are the interview tips for Find Champion I?

Clearly explain your thought process as you iterate through the teams. Be mindful of edge cases, such as the smallest matrix size. Practice explaining your solution in simple terms, as if teaching someone else.

What are common mistakes when solving Find Champion I?

Not considering all teams when updating the champion. Assuming the first team is always the champion without validation.

#2923

Find Champion I

Easy

ArrayMatrixArrayMatrix

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

We can find the champion in a single pass by leveraging the fact that if a team is stronger than another, it can only be the champion if no other team beats it. We can keep track of the potential champion and validate it in one go.

⚙️

Algorithm

5 steps

1Step 1: Initialize a variable `champion` to 0.
2Step 2: Iterate through each team `i` from 1 to n-1.
3Step 3: If grid[champion][i] == 1, it means the current champion is stronger than team i, so continue.
4Step 4: If grid[champion][i] == 0, it means team i is stronger than the current champion, so update `champion` to `i`.
5Step 5: After the loop, return the `champion`.

solution.py8 lines

1# Full working Python code
2
3def findChampion(grid):
4    champion = 0
5    for i in range(1, len(grid)):
6        if grid[champion][i] == 0:
7            champion = i
8    return champion

ℹ

Complexity note: This complexity is optimal because we only make a single pass through the teams, resulting in linear time complexity.

1The champion must be stronger than all other teams, which allows us to eliminate teams quickly.
2The properties of the matrix (0s and 1s) help us determine relationships without needing to check every possible pair.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.