<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> v) { 
	  int cnt1 = 0; int cnt2 = 0; 
	  int ele1 = INT_MIN; int ele2 = INT_MIN; 
	  int n = v.size();
	   for(int i = 0;i<n;i++){
		    if(cnt1 == 0 && ele2 != v[i]){
				 cnt1 = 1; ele1 = v[i]; } 
		    else if(cnt2 == 0 && ele1 != v[i]){
				 cnt2 = 1; ele2 = v[i]; }
		 else if(v[i] == ele1) cnt1++;
		  else if(v[i] == ele2) cnt2++;
		   else { cnt1--; cnt2--; } 
		   }
		    vector<int> ans;
			 cnt1 = 0,cnt2 = 0;
			  for(int i = 0;i<n;i++){
				   if (v[i] == ele1) cnt1++;
				    if (v[i] == ele2) cnt2++;
					 } 
			int mini = (n/3) + 1;
                        if (cnt1 >= mini) {
                                    ans.push_back(ele1);
                        }
                        if (cnt2 >= mini) {
                                    ans.push_back(ele2);
                        }
                        return ans;
					 
    }
};
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