Approach: In-degree count

the champion is the only team with in-degree 0; if there are zero or several such teams, return -1

Complexity: O(n + e) time, O(n) space.

Full Solution — all languages

class Solution:
    def findChampion(self, n, edges):
        indeg = [0] * n
        for a, b in edges:
            indeg[b] += 1
        champ, cnt = -1, 0
        for i in range(n):
            if indeg[i] == 0:
                champ = i
                cnt += 1
        return champ if cnt == 1 else -1

import java.util.*;
class Solution {
    public int findChampion(int n, int[][] edges) {
        int[] indeg = new int[n];
        for (int[] e : edges) indeg[e[1]]++;
        int champ = -1, cnt = 0;
        for (int i = 0; i < n; i++) if (indeg[i] == 0) {
            champ = i;
            cnt++;
        }
        return cnt == 1 ? champ : -1;
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: int findChampion(int n, vector<vector<int>>& edges) {
        vector<int> indeg(n, 0);
        for (auto& e : edges) indeg[e[1]]++;
        int champ = -1, cnt = 0;
        for (int i = 0; i < n; i++) if (indeg[i] == 0) {
            champ = i;
            cnt++;
        }
        return cnt == 1 ? champ : -1;
    }
};

/**
 * @param {number} n
 * @param {number[][]} edges
 * @return {number}
 */ var findChampion = function(n, edges) {
    const indeg = new Array(n).fill(0);
    for (const [a, b] of edges) indeg[b]++;
    let champ = -1, cnt = 0;
    for (let i = 0; i < n; i++) if (indeg[i] === 0) {
        champ = i;
        cnt++;
    }
    return cnt === 1 ? champ : -1;
};

Find Champion II

Example 1

Example 2

Constraints