<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>ans;
        int cnt1=0,cnt2=0;
        int el1=INT_MIN;
        int el2=INT_MIN;
        int n=nums.size();
        for(int i=0;i<n;i++){
            if(cnt1==0 && el2!=nums[i]){
                cnt1=1;
                el1=nums[i];
            }
            else if(cnt2==0 && el1!=nums[i]){
                cnt2=1;
                el2=nums[i];
            }
            else if(nums[i]==el1)cnt1++;
            else if(nums[i]==el2)cnt2++;
            else cnt1--,cnt2--;
        }
        cnt1=0,cnt2=0;
        for(int i=0;i<n;i++){
            if(nums[i]==el1)cnt1++;
            if(nums[i]==el2)cnt2++;
        }
        int check=int(n/3)+1;
        if(cnt1>=check)ans.push_back(el1);
        if(cnt2>=check)ans.push_back(el2);
        return ans;
    }
};
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