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