class Solution: def majorityElement (self, nums: List[int]) -> int: 2 if len(nums) % 2 == 1: count = nums.count(nums [0]) if count > len(nums) // 2: return nums[0] else: 3 5 6. 7 8 nums.pop(0) temp =...

For the second picture Can you fix the loop for the algorithm in python

Extracted text: class Solution: def majorityElement (self, nums: List[int]) -> int: 2 if len(nums) % 2 == 1: count = nums.count(nums [0]) if count > len(nums) // 2: return nums[0] else: 3 5 6. 7 8 nums.pop(0) temp = [] for i in range(0, len(nums), 2): if nums[i] == nums[i + 1]: 10 11 - 12 temp.append(nums [i]) x = self.majorityElement(temp) if nums.count(x) > len(nums) // 2: 13 14 15 return x 16 else: 17 return False


Extracted text: ull T-Mobile 11:21 PM 22% E LeetCode * Pick One 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 = Output: 2 [2,2,1,1,1,2,2] AA leetcode.com