<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 getRotationAngle(vector<int> a, vector<int> b, vector<int> c){
        //(x2-x1)(y3-y1) - (y2-y1)(x3 - x1) cross product of ab and ac vector
        return ((b[0] - a[0]) * (c[1] - a[1])) - ((b[1] - a[1]) * (c[0] - a[0]));
    }
    vector<vector<int>> outerTrees(vector<vector<int>>& trees) {
        //by using convex Hull algorithm
        if(trees.size() <= 3) return trees; // if less than equal to 3 then all need to include
        sort(trees.begin(), trees.end()); // sort them in lexo manner(according to order of x coordinates);
        vector<vector<int>> upperHull; //creating upper hull
        //pushing two points in upper hUll as 'a' and 'b';
        upperHull.push_back(trees[0]);
        upperHull.push_back(trees[1]);
        for(int i=2;i<trees.size();i++){ // considering it as 'c' start traversing from index 2
            int Uhs =upperHull.size();
            //slop / cross product should be greater than 0 counter clock wise angle remove b and add c
            while(upperHull.size() >= 2 && getRotationAngle(upperHull[Uhs - 2], upperHull[Uhs - 1], trees[i]) > 0){
                upperHull.pop_back();
                Uhs--;
            }
            upperHull.push_back(trees[i]);
        }
        reverse(trees.begin(), trees.end());
        vector<vector<int>> lowerHull; //creating lower hull
        //same steps for lower hull as upper hull but reversed array of trees
        lowerHull.push_back(trees[0]);
        lowerHull.push_back(trees[1]);
        for(int i=2;i<trees.size();i++){ // considering it as 'c' start traversing from index 2
            int Lhs =lowerHull.size();
            //slop / cross product should be greater than 0 counter clock wise angle remove b and add c
            while(lowerHull.size() >= 2 && getRotationAngle(lowerHull[Lhs - 2], lowerHull[Lhs - 1], trees[i]) > 0){
                lowerHull.pop_back();
                Lhs--;
            }
            lowerHull.push_back(trees[i]);
        }
        vector <vector<int>> ans;
        map<vector<int>, int> duplicate;
        for(auto it : upperHull) {
            if(duplicate[it] == 1) continue;
            duplicate[it] = 1;
            ans.push_back(it);
        }
        for(auto it : lowerHull) {
            if(duplicate[it] == 1) continue;
            duplicate[it] = 1;
            ans.push_back(it);
        }
        sort(ans.begin(), ans.end());
        return ans;
    }
};

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