<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>

#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
template <class T> int sgn(T x) { return (x > 0) - (x < 0); }
template<class T>
struct Point {
typedef Point P;
T x, y;
explicit Point(T x=0, T y=0) : x(x), y(y) {}
bool operator<(P p) const { return tie(x,y) < tie(p.x,p.y); }
bool operator==(P p) const { return tie(x,y)==tie(p.x,p.y); }
P operator+(P p) const { return P(x+p.x, y+p.y); }
P operator-(P p) const { return P(x-p.x, y-p.y); }
P operator*(T d) const { return P(x*d, y*d); }
P operator/(T d) const { return P(x/d, y/d); }
T dot(P p) const { return x*p.x + y*p.y; }
T cross(P p) const { return x*p.y - y*p.x; }
T cross(P a, P b) const { return (a-*this).cross(b-*this); }
T dist2() const { return x*x + y*y; }
double dist() const { return sqrt((double)dist2()); }
// angle to x−axis in interva l [−pi , pi ]
double angle() const { return atan2(y, x); }
P unit() const { return *this/dist(); } // makes d is t ()=1
P perp() const { return P(-y, x); } // rotates +90 degrees
P normal() const { return perp().unit(); }
// returns point rotated ’a ’ radians ccw around the origin
P rotate(double a) const {
return P(x*cos(a)-y*sin(a),x*sin(a)+y*cos(a)); }
friend ostream& operator<<(ostream& os, P p) {
return os << "(" << p.x << "," << p.y << ")"; }
};
typedef Point<ll> P;
vector<P> convexHull(vector<P> pts) {
if (sz(pts) <= 1) return pts;
sort(all(pts));
vector<P> h(sz(pts)+1);
int s = 0, t = 0;
for (int it = 2; it--; s = --t, reverse(all(pts)))
for (P p : pts) {
while (t >= s + 2 && h[t-2].cross(h[t-1], p) < 0) t--;
if (t >= h.size()) h.push_back(p), t++; else
h[t++] = p;
}
return {h.begin(), h.begin() + t - (t == 2 && h[0] == h[1])};
}

class Solution {
public:
    vector<vector<int>> outerTrees(vector<vector<int>>& trees) {
        vector<P> a;
        for (auto& t : trees) {
            a.push_back(P(t[0], t[1]));
        }
        auto b = convexHull(a);
        set<vector<int>> ans;
        for (auto& p : b) {
            ans.insert({(int)p.x, (int)p.y});
        }
        return vector(ans.begin(), ans.end());
    }
};
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