LeetCode #2923 — EASY

Find Champion I

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

There are n teams numbered from 0 to n - 1 in a tournament.

Given a 0-indexed 2D boolean matrix grid of size n * n. For all i, j that 0 <= i, j <= n - 1 and i != j team i is stronger than team j if grid[i][j] == 1, otherwise, team j is stronger than team i.

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.

Example 1:

Input: grid = [[0,1],[0,0]]
Output: 0
Explanation: 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
Explanation: 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.

Step 01

Brute Force Baseline

Problem summary: There are n teams numbered from 0 to n - 1 in a tournament. Given a 0-indexed 2D boolean matrix grid of size n * n. For all i, j that 0 <= i, j <= n - 1 and i != j team i is stronger than team j if grid[i][j] == 1, otherwise, team j is stronger than team i. 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[[0,1],[0,0]]

Example 2

[[0,0,1],[1,0,1],[0,0,0]]

Step 02

Core Insight

What unlocks the optimal approach

The champion should be stronger than all the other teams.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2923: Find Champion I
class Solution {
    public int findChampion(int[][] grid) {
        int n = grid.length;
        for (int i = 0;; ++i) {
            int cnt = 0;
            for (int j = 0; j < n; ++j) {
                if (i != j && grid[i][j] == 1) {
                    ++cnt;
                }
            }
            if (cnt == n - 1) {
                return i;
            }
        }
    }
}

// Accepted solution for LeetCode #2923: Find Champion I
func findChampion(grid [][]int) int {
	n := len(grid)
	for i := 0; ; i++ {
		cnt := 0
		for j, x := range grid[i] {
			if i != j && x == 1 {
				cnt++
			}
		}
		if cnt == n-1 {
			return i
		}
	}
}

# Accepted solution for LeetCode #2923: Find Champion I
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

// Accepted solution for LeetCode #2923: Find Champion I
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2923: Find Champion I
// class Solution {
//     public int findChampion(int[][] grid) {
//         int n = grid.length;
//         for (int i = 0;; ++i) {
//             int cnt = 0;
//             for (int j = 0; j < n; ++j) {
//                 if (i != j && grid[i][j] == 1) {
//                     ++cnt;
//                 }
//             }
//             if (cnt == n - 1) {
//                 return i;
//             }
//         }
//     }
// }

// Accepted solution for LeetCode #2923: Find Champion I
function findChampion(grid: number[][]): number {
    for (let i = 0, n = grid.length; ; ++i) {
        let cnt = 0;
        for (let j = 0; j < n; ++j) {
            if (i !== j && grid[i][j] === 1) {
                ++cnt;
            }
        }
        if (cnt === n - 1) {
            return i;
        }
    }
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.