<p>Given an array <code>nums</code> of size <code>n</code>, return <em>the majority element</em>.</p>

<p>The majority element is the element that appears more than <code>&lfloor;n / 2&rfloor;</code> times. You may assume that the majority element always exists in the array.</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 = [2,2,1,1,1,2,2]
<strong>Output:</strong> 2
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>1 &lt;= n &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>
<strong>Follow-up:</strong> Could you solve the problem in linear time and in <code>O(1)</code> space?
class Solution:
    def majorityElement(self, nums: List[int]) -> int:
        counter = {num:0 for num in nums}
        n = len(nums)
        for num in nums:
            counter[num] += 1
            if counter[num] > n / 2:
                return num

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