<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 candidate1 = 0, candidate2 = 0, count1 = 0, count2 = 0;
    int n = nums.size();

    // First pass: find potential candidates
    for (int num : nums) {
        if (num == candidate1) {
            count1++;
        } else if (num == candidate2) {
            count2++;
        } else if (count1 == 0) {
            candidate1 = num;
            count1 = 1;
        } else if (count2 == 0) {
            candidate2 = num;
            count2 = 1;
        } else {
            count1--;
            count2--;
        }
    }

    // Second pass: confirm the candidates
    count1 = count2 = 0;
    for (int num : nums) {
        if (num == candidate1) count1++;
        else if (num == candidate2) count2++;
    }

    vector<int> result;
    if (count1 > n / 3) result.push_back(candidate1);
    if (count2 > n / 3) result.push_back(candidate2);

    return result;
}

};
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