169. Majority Element

169. Majority Element

Easy

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:
    def majorityElement(self, nums: List[int]) -> int:
        # c = Counter(nums)
        # return c.most_common()[0][0]

        count = 0
        candidate = None
        for num in nums:
            if count == 0:
                candidate = num
            count += (1 if num == candidate else -1)
        return candidate

Random Note

From python 3.7 dict guarantees that order will be kept as they inserted, and popitem will use LIFO order but we need FIFO type system. so we need OrderedDict which have popIten(last = T/F) for this req. One thing, next(iter(dict)) will return the first key of the dict

Python