LeetCode #997 — EASY

Find the Town Judge

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

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] = [a_i, b_i] representing that the person labeled a_i trusts the person labeled b_i. 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.
a_i != b_i
1 <= a_i, b_i <= n

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

2
[[1,2]]

Example 2

3
[[1,3],[2,3]]

Example 3

3
[[1,3],[2,3],[3,1]]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #997: Find the Town Judge
class Solution {
    public int findJudge(int n, int[][] trust) {
        int[] cnt1 = new int[n + 1];
        int[] cnt2 = new int[n + 1];
        for (var t : trust) {
            int a = t[0], b = t[1];
            ++cnt1[a];
            ++cnt2[b];
        }
        for (int i = 1; i <= n; ++i) {
            if (cnt1[i] == 0 && cnt2[i] == n - 1) {
                return i;
            }
        }
        return -1;
    }
}

// Accepted solution for LeetCode #997: Find the Town Judge
func findJudge(n int, trust [][]int) int {
	cnt1 := make([]int, n+1)
	cnt2 := make([]int, n+1)
	for _, t := range trust {
		a, b := t[0], t[1]
		cnt1[a]++
		cnt2[b]++
	}
	for i := 1; i <= n; i++ {
		if cnt1[i] == 0 && cnt2[i] == n-1 {
			return i
		}
	}
	return -1
}

# Accepted solution for LeetCode #997: Find the Town Judge
class Solution:
    def findJudge(self, n: int, trust: List[List[int]]) -> int:
        cnt1 = [0] * (n + 1)
        cnt2 = [0] * (n + 1)
        for a, b in trust:
            cnt1[a] += 1
            cnt2[b] += 1
        for i in range(1, n + 1):
            if cnt1[i] == 0 and cnt2[i] == n - 1:
                return i
        return -1

// Accepted solution for LeetCode #997: Find the Town Judge
impl Solution {
    pub fn find_judge(n: i32, trust: Vec<Vec<i32>>) -> i32 {
        let mut cnt1 = vec![0; (n + 1) as usize];
        let mut cnt2 = vec![0; (n + 1) as usize];

        for t in trust.iter() {
            let a = t[0] as usize;
            let b = t[1] as usize;
            cnt1[a] += 1;
            cnt2[b] += 1;
        }

        for i in 1..=n as usize {
            if cnt1[i] == 0 && cnt2[i] == (n as usize) - 1 {
                return i as i32;
            }
        }

        -1
    }
}

// Accepted solution for LeetCode #997: Find the Town Judge
function findJudge(n: number, trust: number[][]): number {
    const cnt1: number[] = new Array(n + 1).fill(0);
    const cnt2: number[] = new Array(n + 1).fill(0);
    for (const [a, b] of trust) {
        ++cnt1[a];
        ++cnt2[b];
    }
    for (let i = 1; i <= n; ++i) {
        if (cnt1[i] === 0 && cnt2[i] === n - 1) {
            return i;
        }
    }
    return -1;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.