1000 - 1099
0997 - Find the Town Judge (Easy)
Problem Link
https://leetcode.com/problems/find-the-town-judge/
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.
Return the label of the town judge if the town judge exists and can be identified, or return-1otherwise.
Example 1:
Input: n = 2, trust = [[1,2]]
Output: 2Example 2:
Input: n = 3, trust = [[1,3],[2,3]]
Output: 3Example 3:
Input: n = 3, trust = [[1,3],[2,3],[3,1]]
Output: -1Constraints:
1 <= n <= 10000 <= trust.length <= 10^4trust[i].length == 2- All the pairs of
trustare unique. ai != bi1 <= ai, bi <= n
Approach 1: Counting (Two Arrays)
class Solution {
public:
int findJudge(int n, vector<vector<int>>& trust) {
// indegree (in) = trusted by others
// outdegree (out) = trust others
vector<int> in(n + 1), out(n + 1);
// calculate the in and out
for (auto x : trust) in[x[1]]++, out[x[0]]++;
// iterate each person
for (int i = 1; i <= n; i++) {
// the judge will be the one with indegree = n - 1 and outdegree = 0
if (in[i] == n - 1 && out[i] == 0) {
// found the judge
return i;
}
}
// cannot identify -> return -1
return -1;
}
};
Approach 2: Counting (One Array)
We can further simplify Approach 1 with one array since we know indegrees - outdegrees = n - 1.
class Solution {
public:
int findJudge(int n, vector<vector<int>>& trust) {
// n people
vector<int> v(n + 1);
for (auto x : trust) {
// out: trust other
v[x[0]]--;
// in: trusted by others
v[x[1]]++;
}
for (int i = 1; i <= n; i++) {
// we are looking for indegrees - outdegrees = n - 1
if (v[i] == n - 1) {
// if so, the i-th person will be the judge
return i;
}
}
// cannot identify -> return -1
return -1;
}
};