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
-231 <= nums[i] <= 231 - 1
Follow-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