09 Sept 2026•Go / Python / TypeScript•Easy

Find the Town Judge

Find the person trusted by everyone else who trusts nobody.

View original LeetCode question

Give each person one score for incoming trust and minus one for outgoing trust; the judge's final score is n minus one.

complexity

O(n + t) time and O(n) space.

solution files

Go find-the-town-judge/solution.go
Python find-the-town-judge/solution.py
TypeScript find-the-town-judge/solution.ts

Solution files

Go•find-the-town-judge/solution.go

package main

func findJudge(n int, trust [][]int) int {
	score := make([]int, n+1)
	for _, relation := range trust {
		score[relation[0]]--
		score[relation[1]]++
	}
	for person := 1; person <= n; person++ {
		if score[person] == n-1 {
			return person
		}
	}
	return -1
}

Python•find-the-town-judge/solution.py

class Solution:
    def findJudge(self, n: int, trust: list[list[int]]) -> int:
        score = [class="syntax-number">0] * (n + class="syntax-number">1)
        for source, target in trust: score[source] -= class="syntax-number">1; score[target] += class="syntax-number">1
        return next((person for person in range(class="syntax-number">1, n + class="syntax-number">1) if score[person] == n - class="syntax-number">1), -class="syntax-number">1)

TypeScript•find-the-town-judge/solution.ts

function findJudge(n: number, trust: number[][]): number {
  const score = new Array<number>(n + class="syntax-number">1).fill(class="syntax-number">0);
  for (const [from, to] of trust) { score[from] -= class="syntax-number">1; score[to] += class="syntax-number">1; }
  for (let person = class="syntax-number">1; person <= n; person += class="syntax-number">1) if (score[person] === n - class="syntax-number">1) return person;
  return -class="syntax-number">1;
}