<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) {
        if(nums.size()==1) return nums;
        int x,y;
        x=INT_MAX; y=INT_MAX;
        int count_x, count_y;
        count_x=0; count_y=0;
        for(int i=0; i<nums.size(); i++){
            if(x!=nums[i] && y!=nums[i]){
                if(count_x==0){
                    x=nums[i];
                    count_x++;
                }
                else if(count_y==0){
                    y=nums[i];
                    count_y++;
                }
                else{
                    count_x--;
                    count_y--;
                }
            }
            else if(x==nums[i]) count_x++;
            else if(y==nums[i]) count_y++;
        }
        count_x=0; count_y=0;
        for(int it: nums){
            if(it==x) count_x++;
            else if(it==y) count_y++;
        }
        if(count_x>nums.size()/3 && count_y>nums.size()/3) return {x, y};
        else if(count_x>nums.size()/3) return {x};
        else if(count_y>nums.size()/3) return {y};
        else return {};
    }
};
18800