<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>

class Solution {
public:
    vector<int> majorityElement(vector<int>& nums) 
    {
        vector <int> v;
        for(int i=0;i<nums.size();i++)
        {
            int count=0;
            
            for(int j=0;j<nums.size();j++)
            {
                if(nums[i]==nums[j])
                {
                    count++;
                }
            }
            if(count>(nums.size())/3)
            {
                
                v.push_back(nums[i]);
                
            }
        }
        for(int i=0;i<v.size();i++)
        {
            for(int j=i+1;j<v.size();j++)
            {
                if(v[i]==v[j])
                {
                    v.erase(v.begin()+j);
                    j--;
                }
            }
        }
        return v;
        
    }
};
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