<p>You are given an array <code>trees</code> where <code>trees[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> represents the location of a tree in the garden.</p>

<p>Fence the entire garden using the minimum length of rope, as it is expensive. The garden is well-fenced only if <strong>all the trees are enclosed</strong>.</p>

<p>Return <em>the coordinates of trees that are exactly located on the fence perimeter</em>. You may return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/24/erect2-plane.jpg" style="width: 400px; height: 393px;" />
<pre>
<strong>Input:</strong> trees = [[1,1],[2,2],[2,0],[2,4],[3,3],[4,2]]
<strong>Output:</strong> [[1,1],[2,0],[4,2],[3,3],[2,4]]
<strong>Explanation:</strong> All the trees will be on the perimeter of the fence except the tree at [2, 2], which will be inside the fence.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/24/erect1-plane.jpg" style="width: 400px; height: 393px;" />
<pre>
<strong>Input:</strong> trees = [[1,2],[2,2],[4,2]]
<strong>Output:</strong> [[4,2],[2,2],[1,2]]
<strong>Explanation:</strong> The fence forms a line that passes through all the trees.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= trees.length &lt;= 3000</code></li>
	<li><code>trees[i].length == 2</code></li>
	<li><code>0 &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= 100</code></li>
	<li>All the given positions are <strong>unique</strong>.</li>
</ul>

class Solution {
public:
    vector<vector<int>> outerTrees(vector<vector<int>>& trees) {
        if (trees.size() <= 3) return trees;
        
        // Sorting trees by x and y coordinates
        sort(trees.begin(), trees.end(), [](const vector<int>& a, const vector<int>& b) {
            return a[0] == b[0] ? a[1] < b[1] : a[0] < b[0];
        });

        // Helper function to calculate the cross product of vectors OA and OB
        // Positive result means counterclockwise, 0 means collinear, negative means clockwise
        auto cross = [](const vector<int>& O, const vector<int>& A, const vector<int>& B) {
            return (A[0] - O[0]) * (B[1] - O[1]) - (A[1] - O[1]) * (B[0] - O[0]);
        };

        vector<vector<int>> hull;

        // Building the lower hull
        for (const auto& tree : trees) {
            while (hull.size() >= 2 && cross(hull[hull.size() - 2], hull.back(), tree) < 0) {
                hull.pop_back();
            }
            hull.push_back(tree);
        }

        // Building the upper hull
        int t = hull.size() + 1;
        for (int i = trees.size() - 1; i >= 0; --i) {
            while (hull.size() >= t && cross(hull[hull.size() - 2], hull.back(), trees[i]) < 0) {
                hull.pop_back();
            }
            hull.push_back(trees[i]);
        }

        // Removing duplicates
        hull.pop_back();
        sort(hull.begin(), hull.end());
        hull.erase(unique(hull.begin(), hull.end()), hull.end());

        return hull;
    }
};

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