Skip to main content

23. Majority Element

easyAsked at Reddit

Find the element that appears more than n/2 times in an array. Reddit uses Boyer-Moore voting to test whether candidates recognize specialized algorithms — the same vote-majority pattern matters for their vote-fraud detection (one IP shouldn't dominate a thread).

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Source citations

Public interview reports confirming this problem appears in Reddit loops.

  • Glassdoor (2026-Q1)— Reddit trust-and-safety phone screen — followed by vote-fraud detection follow-ups.
  • Blind (2025-10)— Reported as a Reddit fraud-team favorite.

Problem

Given an array nums of size n, return the majority element. The majority element is the element that appears more than n/2 times. You may assume that the majority element always exists in the array. Follow-up: Could you solve the problem in linear time and in O(1) space?

Constraints

  • n == nums.length
  • 1 <= n <= 5 * 10^4
  • -10^9 <= nums[i] <= 10^9

Examples

Example 1

Input
nums = [3,2,3]
Output
3

Example 2

Input
nums = [2,2,1,1,1,2,2]
Output
2

Approaches

1. Hash count

Count each element. Return the one with count > n/2.

Time
O(n)
Space
O(n)
function majorityElement(nums) {
  const count = new Map();
  for (const n of nums) {
    count.set(n, (count.get(n) || 0) + 1);
    if (count.get(n) > nums.length / 2) return n;
  }
}

Tradeoff: Linear time but O(n) space — fails the follow-up.

2. Boyer-Moore voting (optimal)

Single pass with a candidate and count. On match, count++. On mismatch, count--. When count hits 0, switch candidate.

Time
O(n)
Space
O(1)
function majorityElement(nums) {
  let candidate = null, count = 0;
  for (const n of nums) {
    if (count === 0) candidate = n;
    count += (n === candidate) ? 1 : -1;
  }
  return candidate;
}

Tradeoff: O(1) space. Beautiful algorithm — works because a true majority survives all cancellations.

Reddit-specific tips

Reddit interviewers want Boyer-Moore named explicitly. Bonus signal: connect this to their bot-detection — if a single user/IP appears in majority of votes on a post, the vote is likely brigaded. The intuition is the same: a single dominant signal survives cancellations.

Common mistakes

  • Comparing count > n/2 inside the loop (only valid if majority is guaranteed).
  • Forgetting to reassign candidate when count hits 0.
  • Initializing count to 1 (must start at 0 so the first element becomes candidate).

Follow-up questions

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

  • Majority element II (LC 229): elements appearing > n/3 times. Two candidates.
  • What if no guarantee of majority? Verify with a second pass.
  • Distributed majority — combining counts across shards.

Solve it now

Free. No sign-up. Python and JavaScript run instantly in your browser.

Output

Press Run or Cmd+Enter to execute

FAQ

Why does Boyer-Moore work?

Pair each majority element with a non-majority one. Since majority is > n/2, there's always one unpaired — the survivor.

What if majority isn't guaranteed?

Run a second pass to verify the candidate's count is actually > n/2.

Practice these live with InterviewChamp.AI

Drill Majority Element and other Reddit interview questions under real-loop conditions with instant feedback on your reasoning, complexity claims, and code.

Practice these live with InterviewChamp.AI →