<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 size = nums.size();
        if (size == 0) {
            return vector<int>{0};
        } 
        map<int, int> m;
        int num = size / 3;
        if(size == 1 && nums[0] > num) {
            return nums;
        }

        set <int> v;
        for (auto it : nums) {
            if (m.contains(it)) {
                m[it] += 1;
            } else {
                m[it] = 1;
                // if(1 > num) {
                // v.push_back(it);
                // }
            } 
        if (m[it] > num) {
            v.insert(it);
        }
        }
    // cout<<"\n";
    // unordered_set <int> s(v.begin(), v.end());
    // v.clear();
    // copy(s.begin(), s.end(), back_inserter(v));

    return vector <int> (v.begin(), v.end());
    }
};
19000