<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;

        int n = trees.size();
        vector<vector<int>> top, bottom, middle;

        sort(trees.begin(), trees.end(), [](const std::vector<int>& a, const std::vector<int>& b) {
            return (a[0] == b[0]) ? (a[1] < b[1]) : (a[0] < b[0]);
        });

        vector<int> start = trees[0], end = trees[n-1];

        float slope = (double)(end[1] - start[1]) / (end[0] - start[0]);

        top.push_back(start);
        bottom.push_back(start);

        for (int i = 1; i < n-1; i++) {
            auto p = trees[i];
            double v = start[1]+ (p[0]-start[0])*slope;

            if (p[1] == v) middle.push_back(p);
            else if (p[1] > v) top.push_back(p);
            else bottom.push_back(p);
        }

        top.push_back(end);
        bottom.push_back(end);

        // Try to backwards evaluate top perimeter
        for (int i = 2; i < top.size(); i++) {
            for (int j = i-1; j > 0; j--) {
                double i_slope = (double)(top[i][1] - top[i-2][1]) / (top[i][0] - top[i-2][0]);
                double j_slope = (double)(top[j][1] - top[j-1][1]) / (top[j][0] - top[j-1][0]);

                if (i_slope <= j_slope) break;

                top.erase(top.begin()+j, top.begin()+j+1);
                i--;
            }
        }

        // Try to backwards evaluate bottom perimeter
        for (int i = 2; i < bottom.size(); i++) {
            for (int j = i-1; j > 0; j--) {
                double i_slope = (double)(bottom[i][1] - bottom[i-2][1]) / (bottom[i][0] - bottom[i-2][0]);
                double j_slope = (double)(bottom[j][1] - bottom[j-1][1]) / (bottom[j][0] - bottom[j-1][0]);

                if (i_slope >= j_slope) break;

                bottom.erase(bottom.begin()+j, bottom.begin()+j+1);
                i--;
            }
        }

        vector<vector<int>> output = {start, end};

        output.insert(output.begin(), top.begin()+1, top.end()-1);
        output.insert(output.begin(), bottom.begin()+1, bottom.end()-1);
        if (top.size() == 2 || bottom.size() == 2) output.insert(output.begin(), middle.begin(), middle.end());

        return output;
    }
};
30000