<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 n = v.size(); //size of the array

    int cnt1 = 0, cnt2 = 0; // counts
    int el1 = INT_MIN; // element 1
    int el2 = INT_MIN; // element 2

    // applying the Extended Boyer Moore's Voting Algorithm:
    for (int i = 0; i < n; i++) {
        if (cnt1 == 0 && el2 != v[i]) {
            cnt1 = 1;
            el1 = v[i];
        }
        else if (cnt2 == 0 && el1 != v[i]) {
            cnt2 = 1;
            el2 = v[i];
        }
        else if (v[i] == el1) cnt1++;
        else if (v[i] == el2) cnt2++;
        else {
            cnt1--, cnt2--;
        }
    }

    vector<int> ls; // list of answers

    // Manually check if the stored elements in
    // el1 and el2 are the majority elements:
    cnt1 = 0, cnt2 = 0;
    for (int i = 0; i < n; i++) {
        if (v[i] == el1) cnt1++;
        if (v[i] == el2) cnt2++;
    }

    int mini = int(n / 3) + 1;
    if (cnt1 >= mini) ls.push_back(el1);
    if (cnt2 >= mini) ls.push_back(el2);

    // Uncomment the following line
    // if it is told to sort the answer array:
    // sort(ls.begin(), ls.end()); //TC --> O(2*log2) ~ O(1);

    return ls;
    }
};
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