In this guide, you will get 997. Find the Town Judge LeetCode Solution with the best time and space complexity. The solution to Find the Town Judge problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
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
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 {
public:
int findJudge(int n, vector<vector<int>>& trust) {
vector<int> count(n + 1);
for (vector<int>& t : trust) {
--count[t[0]];
++count[t[1]];
}
for (int i = 1; i < n + 1; ++i)
if (count[i] == n - 1)
return i;
return -1;
}
}; /* code provided by PROGIEZ */
997. Find the Town Judge LeetCode Solution in Java
class Solution {
public int findJudge(int n, int[][] trust) {
int[] count = new int[n + 1];
for (int[] t : trust) {
--count[t[0]];
++count[t[1]];
}
for (int i = 1; i < n + 1; ++i)
if (count[i] == n - 1)
return i;
return -1;
}
} // code provided by PROGIEZ
997. Find the Town Judge LeetCode Solution in Python
class Solution:
def findJudge(self, n: int, trust: list[list[int]]) -> int:
count = [0] * (n + 1)
for a, b in trust:
count[a] -= 1
count[b] += 1
for i in range(1, n + 1):
if count[i] == n - 1:
return i
return -1 # code by PROGIEZ
