In a town, there are n
people labeled from 1
to n
. There is a rumor that one of these people is secretly the town judge.
If the town judge exists, then:
- The town judge trusts nobody.
- Everybody (except for the town judge) trusts the town judge.
- There is exactly one person that satisfies properties 1 and 2.
You are given an array trust
where trust[i] = [ai, bi]
representing that the person labeled ai
trusts the person labeled bi
.
Return the label of the town judge if the town judge exists and can be identified, or return -1
otherwise.
Example 1:
Input: n = 2, trust = [[1,2]] Output: 2
Example 2:
Input: n = 3, trust = [[1,3],[2,3]] Output: 3
Example 3:
Input: n = 3, trust = [[1,3],[2,3],[3,1]] Output: -1
Constraints:
1 <= n <= 1000
0 <= trust.length <= 104
trust[i].length == 2
- All the pairs of
trust
are unique. ai != bi
1 <= ai, bi <= n
class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
N = 1001
c1, c2 = [0] * N, [0] * N
for a, b in trust:
c1[a] += 1
c2[b] += 1
for i in range(1, N):
if c1[i] == 0 and c2[i] == n - 1:
return i
return -1
class Solution {
public int findJudge(int n, int[][] trust) {
int N = 1001;
int[] c1 = new int[N];
int[] c2 = new int[N];
for (int[] e : trust) {
++c1[e[0]];
++c2[e[1]];
}
for (int i = 1; i < N; ++i) {
if (c1[i] == 0 && c2[i] == n - 1) {
return i;
}
}
return -1;
}
}
function findJudge(n: number, trust: number[][]): number {
let candidates = new Array(n).fill(0);
for (let [a, b] of trust) {
candidates[a - 1] = -1;
if (candidates[b - 1] >= 0) {
candidates[b - 1]++;
}
}
for (let i = 0; i < n; i++) {
if (candidates[i] == n - 1) {
return i + 1;
}
}
return -1;
}
class Solution {
public:
int findJudge(int n, vector<vector<int>>& trust) {
int N = 1001;
vector<int> c1(N);
vector<int> c2(N);
for (auto& e : trust) {
++c1[e[0]];
++c2[e[1]];
}
for (int i = 1; i < N; ++i) {
if (c1[i] == 0 && c2[i] == n - 1) return i;
}
return -1;
}
};
func findJudge(n int, trust [][]int) int {
N := 1001
c1 := make([]int, N)
c2 := make([]int, N)
for _, e := range trust {
c1[e[0]]++
c2[e[1]]++
}
for i := 1; i < N; i++ {
if c1[i] == 0 && c2[i] == n-1 {
return i
}
}
return -1
}