Problem Statement
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
. If a trust relationship does not exist in trust
array, then such a trust relationship does not exist.
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
Try here before watching the video.
Video Solution
Java Code
class Solution {
public int findJudge(int n, int[][] trust) {
int[] degree = new int[n + 1];
for(int[] trustPair : trust) {
degree[trustPair[0]] -= 1;
degree[trustPair[1]] += 1;
}
for(int i = 1; i <= n; i++) {
if(degree[i] == (n - 1)) {
return i;
}
}
return -1;
}
}
C++ Code
#include <vector>
using namespace std;
class Solution {
public:
int findJudge(int n, vector<vector<int>>& trust) {
vector<int> degree(n + 1);
for (const auto& trustPair : trust) {
degree[trustPair[0]]--;
degree[trustPair[1]]++;
}
for (int i = 1; i <= n; i++) {
if (degree[i] == (n - 1)) {
return i;
}
}
return -1;
}
};
Python Code
class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
degree = [0] * (n + 1)
for trustPair in trust:
degree[trustPair[0]] -= 1
degree[trustPair[1]] += 1
for i in range(1, n + 1):
if degree[i] == n - 1:
return i
return -1
Javascript Code
var findJudge = function(n, trust) {
const degree = new Array(n + 1).fill(0);
for (const trustPair of trust) {
degree[trustPair[0]]--;
degree[trustPair[1]]++;
}
for (let i = 1; i <= n; i++) {
if (degree[i] === n - 1) {
return i;
}
}
return -1;
};
Go Code
func findJudge(n int, trust [][]int) int {
degree := make([]int, n+1)
for _, trustPair := range trust {
degree[trustPair[0]]--
degree[trustPair[1]]++
}
for i := 1; i <= n; i++ {
if degree[i] == n-1 {
return i
}
}
return -1
}
Complexity Analysis
Time Complexity: O(n)
Space Complexity: O(n)