Is Majority Element easy or hard?

Majority Element is classified as an Easy problem on LeetCode with an acceptance rate around 66%. The HashMap solution is straightforward, but recognizing and implementing the Boyer-Moore Voting Algorithm demonstrates deeper algorithm knowledge.

Majority Element Python/Java solution

Python and Java implementations commonly use either a HashMap for counting frequencies or the Boyer-Moore Voting Algorithm. The Boyer-Moore approach is preferred because it runs in O(n) time and O(1) space while requiring only two variables.

How to solve Majority Element in O(n)?

Use the Boyer-Moore Voting Algorithm. Iterate through the array while tracking a candidate and a count. Increase the count when the current element matches the candidate and decrease it otherwise; when the count reaches zero, select a new candidate.

What is the best approach for Majority Element?

The Boyer-Moore Voting Algorithm is the best approach. It finds the majority element in O(n) time using only O(1) extra space. The algorithm maintains a candidate and a counter that cancels out non-majority elements during a single pass through the array.

Is Majority Element asked at Google/Amazon/Meta?

Majority Element is a common interview question across companies like Amazon, Google, Meta, and Microsoft. It tests understanding of array traversal, frequency counting, and the Boyer-Moore Voting Algorithm optimization.

What data structure is used in Majority Element?

The problem can be solved using a hash table for frequency counting or by using constant variables in the Boyer-Moore Voting Algorithm. Both rely on scanning an array and tracking occurrences of elements.

What is the time complexity of Majority Element?

The optimal solution runs in O(n) time. Boyer-Moore Voting scans the array once while maintaining a running candidate. HashMap counting also runs in O(n) time but requires O(n) additional space for storing frequencies.

169. Majority Element - Practice Online

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.

Example 1:

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

Example 2:

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

Constraints:

n == nums.length
1 <= n <= 5 * 10⁴
-10⁹ <= nums[i] <= 10⁹

Follow-up: Could you solve the problem in linear time and in O(1) space?