<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:
static vector<vector<int>> outerTrees(vector<vector<int>>& points) {
        // Andrew's monotone chain method.
        sort(begin(points), end(points), [](const vector<int>& a, const vector<int>& b) {
            return a[0] < b[0] || (a[0] == b[0] && a[1] < b[1]);
        });
        // left to right
        const int n = size(points);
        vector<vector<int>> ans;
        ans.reserve(n);
        for (int i = 0; i < n; ++i) {
            while (size(ans) > 1 && orientation(ans[size(ans) - 2], ans.back(), points[i]) < 0) 
                ans.pop_back();
            ans.push_back(points[i]);
        }
        // If all points are along a line, size(ans) is n after left to right procedure.
        if (size(ans) == n) return ans;

        // right to left
        for (int i = n - 2; i >= 0; --i) {
            while (size(ans) > 1 && orientation(ans[size(ans) - 2], ans.back(), points[i]) < 0) 
                ans.pop_back();
            ans.push_back(points[i]);
        }
        ans.pop_back();
        return ans;
    }

    static int orientation(const vector<int>& a, const vector<int>& b, const vector<int>& c) {
        return (b[0] - a[0]) * (c[1] - b[1]) - (b[1] - a[1]) * (c[0] - b[0]);
    }
};
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