Leetcode 169. Majority Element (Python)

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?

class Solution(object):
    def majorityElement(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        r = []
        for i in nums:
            if i not in r:
                r.append(i)
                
        if len(r) == 2:
            first = nums[0]
            list1 = []
            list2 = []

            for i in nums:
                if first == i:
                    list1.append(i)
                else:
                    list2.append(i)


            if len(list1) < len(list2):
                return list2[0]
            else:
                return first
            
        if len(r) == 3:
            first = nums[0]
            list3 = []
            list4 = []
            list5 = []
            list6 = []

            for i in nums:
                if first == i:
                    list3.append(i)
                else:
                    list4.append(i)
                
            second = list4[0]
            for j in list4:
                if second == j:
                    list5.append(j)
                else:
                    list6.append(j)

            if len(list3) > len(list5):
                if len(list3) > list(list6):
                    return list3[0]
                else:
                    return list6[0]
            else:
                if len(list5) > len(list6):
                    return list5[0]
                else:
                    return list6[0]
            
            
        res = {}
        for i in nums:
            res[i] = nums.count(i)

        inverse = [(value, key) for key, value in res.items()]
        return max(inverse)[1]