<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) {
        int n=nums.size();
        vector<int> ans;
        int c1=0,c2=0,e1,e2;
        for(int i=0;i<n;i++)
        {
            if(c1==0 && e2!=nums[i])
            {
                c1++;
                e1=nums[i];
            }
            else if(c2==0 && e1!=nums[i])
            {
                c2++;
                e2=nums[i];
            }
            else if(e1==nums[i])
            c1++;
            else if(e2==nums[i])
            c2++;
            else
            {
                c1--;
                c2--;
            }
        }

    c1=0,c2=0;
    for(int i=0;i<n;i++)
    {
        if(nums[i]==e1)
        c1++;
        if(nums[i]==e2)
        c2++;
    }

    if(c1>n/3)
    ans.push_back(e1);
    if(c2>n/3 && e2!=e1)
    ans.push_back(e2);

    return ans;
    }
};
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