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997. Find the Town Judge

easyAsked at Cisco

Find the Town Judge at Cisco is a graph in-degree / out-degree problem disguised as a trust puzzle. The interviewer is checking whether you can recast 'judge = everyone trusts them, they trust no one' as 'in-degree N-1, out-degree 0' on a directed graph.

By Sam K., Founder, InterviewChamp.AI · Last verified

Source citations

Public interview reports confirming this problem appears in Cisco loops.

  • Glassdoor (2026-Q1)— Cisco SDE-I and SDE-II reports cite this as a 20-minute warm-up graph round.
  • Reddit r/leetcode (2025-11)— Cisco interview write-up lists Find the Town Judge as a recurring problem.

Problem

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, and 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 uniquely identified, or return -1 otherwise.

Constraints

  • 1 <= n <= 1000
  • 0 <= trust.length <= 10^4
  • trust[i].length == 2
  • All the pairs of trust are unique.
  • ai != bi
  • 1 <= ai, bi <= n

Examples

Example 1

Input
n = 2, trust = [[1,2]]
Output
2

Explanation: 1 trusts 2, 2 trusts no one → 2 is the judge.

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

Explanation: 3 trusts 1, so 3 can't be the judge.

Approaches

1. Two-array brute force

Build separate trustsCount (out-degree) and trustedByCount (in-degree) arrays. The judge has trustedByCount[judge] == n-1 AND trustsCount[judge] == 0.

Time
O(E + N)
Space
O(N)
function findJudgeBrute(n, trust) {
  const trusts = new Array(n + 1).fill(0);
  const trustedBy = new Array(n + 1).fill(0);
  for (const [a, b] of trust) {
    trusts[a]++;
    trustedBy[b]++;
  }
  for (let i = 1; i <= n; i++) {
    if (trusts[i] === 0 && trustedBy[i] === n - 1) return i;
  }
  return -1;
}

Tradeoff: Clear two-step intent — separate counters for in and out degree. Correct, linear time. Slightly more space than necessary.

2. Net-trust single-array (optimal)

Use ONE array `net`. For each edge [a, b]: net[a]--; net[b]++. The judge is the unique node with net[i] == n-1.

Time
O(E + N)
Space
O(N)
function findJudge(n, trust) {
  const net = new Array(n + 1).fill(0);
  for (const [a, b] of trust) {
    net[a]--;
    net[b]++;
  }
  for (let i = 1; i <= n; i++) {
    if (net[i] === n - 1) return i;
  }
  return -1;
}

Tradeoff: Same Big-O, half the memory. The math: judge's net = (n-1) trusted_by - 0 trusts = n-1. Any non-judge has at MOST (n-2) - 1 = n-3 net (trusted by everyone except themselves and the judge, and they trust the judge), so n-1 uniquely identifies the judge.

Cisco-specific tips

Cisco interviewers grade this on whether you recast the problem as 'in-degree N-1, out-degree 0.' Say that sentence first. Then offer the net-trust optimization: 'Since the judge's signature is in - out = n-1, I can collapse to one array.' The math justification — 'no non-judge can have net = n-1 because they trust at least one person AND aren't trusted by themselves' — earns full marks.

Common mistakes

  • Using a 0-indexed array when people are 1-indexed — off-by-one returns -1 on valid inputs.
  • Forgetting the n=1 edge case — trust is empty, person 1 is trivially the judge (n-1 = 0 trustees).
  • Checking only `trustedBy[i] === n-1` and missing the `trusts[i] === 0` half — fails when someone is trusted by all but also trusts someone.

Follow-up questions

An interviewer at Cisco may pivot to one of these next:

  • Find the Celebrity (LC 277 premium) — same in-degree/out-degree idea but only `knows(a, b)` queries available, must minimize queries.
  • What if there can be multiple judges? (Return list of all nodes with net = n-1.)
  • What if trust is weighted? (Now it's about ranking, not single max — sort by weighted in-out.)

Solve it now

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Output

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FAQ

Why does the net = n-1 uniqueness work?

Judge has +n-1 (everyone trusts them) and -0 (they trust no one) → net = n-1. Any non-judge i has at most +n-2 incoming (judge doesn't trust them, themselves doesn't count) and AT LEAST -1 outgoing (they trust the judge if one exists) → net ≤ n-3. So n-1 cleanly identifies the judge.

Is this really a 'graph' problem or just counting?

It IS counting — but the FRAMING as a directed graph with in/out-degree is what makes the algorithm obvious. Cisco interviewers want you to say 'I'll model trust as a directed edge a -> b and the judge is the node with in-degree n-1 and out-degree 0.' That graph framing is the signal.

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