Majority Element II
<p>Given an integer array of size <code>n</code>, find all elements that appear more than <code>&lfloor; n/3 &rfloor;</code> times.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,2,3]
<strong>Output:</strong> [3]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1]
<strong>Output:</strong> [1]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2]
<strong>Output:</strong> [1,2]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Could you solve the problem in linear time and in <code>O(1)</code> space?</p>
Array; Hash Table; Counting; Sorting
class Solution:
    def majorityElement(self, nums: List[int]) -> List[int]:
        n1 = n2 = 0
        m1, m2 = 0, 1
        for m in nums:
            if m == m1:
                n1 += 1
            elif m == m2:
                n2 += 1
            elif n1 == 0:
                m1, n1 = m, 1
            elif n2 == 0:
                m2, n2 = m, 1
            else:
                n1, n2 = n1 - 1, n2 - 1
        return [m for m in [m1, m2] if nums.count(m) > len(nums) // 3]
