Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.

You may assume that the array is non-empty and the majority element always exist in the array.

Example1: Input: [3,2,3] Output: 3

Example2: Input: [2,2,1,1,1,2,2] Output: 2

My solution:

class Solution:
    def majorityElement(self, nums: List[int]) -> int:
        length = len(nums)
        for num in set(nums):
            if nums.count(num) >= (length /2):
                return num
            
# beats 95.6% of submission     
#check each element in set

Other Solutions(Hashmap):⌗

We can use a Hashmap that maps elements to counts in order to count occurrenc in linear time by looping over nums

class Solution:
    def majorityElement(self, nums):
    counts = collections.Counter(nums)
    return max(counts.keys(), key=counts.get)

Thought: Even my submission is faster than 95%, it still takes O(n^2) time. hashmap is obviusly better