[LeetCode] 169. Majority Element

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
• -231 <= nums[i] <= 231 - 1

class Solution:
    def majorityElement(self, nums: List[int]) -> int:
      A  = dict(zip(nums, [nums.count(i) for i in nums]))
      maxVal = max(A.values())
      AReverse = dict(zip([nums.count(i) for i in nums],nums))
      return AReverse[maxVal]
      

# 자연스럽게 Time Limit Exceeded 
# collections library 를 가져와야겠다..

from collections import Counter
class Solution:
    def majorityElement(self, nums: List[int]) -> int:
        A = Counter(nums)
        maxVal = max(A.values())
        AReverse = dict(zip(A.values(),A.keys()))
        return AReverse[maxVal]

#memo 

Couter() 함수 내부코드가 궁금해진다.. 속도차이가.. 

 

Runtime: 152 ms, faster than 98.22% of Python3 online submissions for Majority Element.

Memory Usage: 15.4 MB, less than 96.46% of Python3 online submissions for Majority Element.