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.length1 <= n <= 5 * 104-231 <= nums[i] <= 231 - 1Follow-up: Could you solve the problem in linear time and in O(1) space?
# @param {Integer[]} nums
# @return {Integer}
def majority_element(nums)
  candidate_count = 0
  candidate_letter = nil
  nums.each_with_object({}) do |num, hash|
    hash[num] = 1 + hash[num] ||= 0
  end.each do |k, v|
    next if candidate_count >= v
    candidate_letter = k
    candidate_count = v
  end
  candidate_letter
end