6. Majority Element
easyAsked at RevolutFind the value appearing more than n/2 times, a Revolut warm-up that mirrors detecting the dominant currency in a stream of FX trades.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
Given an array nums of size n, return the majority element which appears more than n/2 times. You may assume the majority element always exists in the array.
Constraints
n == nums.length1 <= n <= 5 * 10^4Majority element always exists
Examples
Example 1
nums=[3,2,3]3Example 2
nums=[2,2,1,1,1,2,2]2Approaches
1. Hash count
Count and return whichever crosses n/2.
- Time
- O(n)
- Space
- O(n)
const m = new Map();
for (const x of nums){ m.set(x,(m.get(x)||0)+1); if (m.get(x)>nums.length/2) return x; }Tradeoff:
2. Boyer-Moore voting
Maintain a candidate and count; non-majority cancel out pairwise. Constant space.
- Time
- O(n)
- Space
- O(1)
function majorityElement(nums){
let cand = 0, count = 0;
for (const x of nums){
if (count === 0) cand = x;
count += (x === cand) ? 1 : -1;
}
return cand;
}Tradeoff:
Revolut-specific tips
Revolut interviewers love Boyer-Moore here because it mirrors streaming FX-pair aggregation — explicitly state the algorithm requires the majority to actually exist or a second verification pass is needed.
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