<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:
    int angle(vector<int> p, vector<int> q, vector<int> r){
        return (q[1]-p[1])*(r[0]-q[0])-(q[0]-p[0])*(r[1]-q[1]);
    }

    vector<vector<int>> outerTrees(vector<vector<int>>& trees) {
        vector<vector<int>> ans;

        sort(trees.begin(), trees.end());

        for(int i=0; i<trees.size(); i++){
            while(ans.size()>1 && angle(ans[ans.size()-2], ans.back(), trees[i])>0){
                ans.pop_back();
            }
            ans.push_back(trees[i]);
        }

        if(ans.size()==trees.size()) return ans;  //no more fence can be allocated

        //now run the entire algo in reverse operation 
        for(int i=trees.size()-2; i>=0; i--){
            while(ans.size()>1 && angle(ans[ans.size()-2], ans.back(), trees[i])>0){
                ans.pop_back();
            }
            ans.push_back(trees[i]);
        }

        //since the starting point will be repeated so pop it
        ans.pop_back();
        return ans;
    }
};
47300