<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) 
    {
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);
        int n = nums.size();
        int maj = n/3;
        
        int maj1 = 0, maj2 = 0;
        
        int vote1 = 0,vote2 = 0;
        for(int i = 0 ;i<n;i++)
        {
            if(nums[i] == maj1)
                vote1++;
            else if(nums[i] == maj2)
                vote2++;
            else if(vote1 == 0)
            {
                maj1 = nums[i];
                vote1 = 1;
            }
            else if(vote2 == 0)
            {
                maj2 = nums[i];
                vote2 = 1;
            }
            else
            {
                vote1--;
                vote2--;
            }
        }
        vote1 = 0;
        vote2 = 0;

        for(int i = 0;i<nums.size();i++)
        {
            if(nums[i] == maj1)
                vote1++;
            else if(nums[i] == maj2)
                vote2++;
        }

        vector<int> ans;
        if(vote1 > maj)
            ans.push_back(maj1);
        if(vote2 > maj)
            ans.push_back(maj2);
        
        return ans;
    }
};
18900