# C++ and Python easy wiki solution

• C++ version:

``````// Ref: http://www.algorithmist.com/index.php/Monotone_Chain_Convex_Hull.cpp
class Solution {
public:
typedef int coord_t;  // coordinate type
typedef long long coord2_t;  // must be big enough to hold 2*max(|coordinate|)^2
// 2D cross product of OA and OB vectors, i.e. z-component of their 3D cross
// product. Returns a positive value, if OAB makes a counter-clockwise turn,
// negative for clockwise turn, and zero if the points are collinear.
coord2_t cross(const Point &O, const Point &A, const Point &B) {
return (A.x - O.x) * (coord2_t)(B.y - O.y) -
(A.y - O.y) * (coord2_t)(B.x - O.x);
}

static bool cmp(Point &p1, Point &p2) {
return p1.x < p2.x || (p1.x == p2.x && p1.y < p2.y);
}

static bool equ(Point &p1, Point &p2) { return p1.x == p2.x && p1.y == p2.y; }
// Returns a list of points on the convex hull in counter-clockwise order.
// Note: the last point in the returned list is the same as the first one.
vector<Point> outerTrees(vector<Point> &P) {
int n = P.size(), k = 0;
vector<Point> H(2 * n);

// Sort points lexicographically
sort(P.begin(), P.end(), cmp);

// Build lower hull
for (int i = 0; i < n; i++) {
while (k >= 2 && cross(H[k - 2], H[k - 1], P[i]) < 0) k--;
H[k++] = P[i];
}

// Build upper hull
for (int i = n - 2, t = k + 1; i >= 0; i--) {
while (k >= t && cross(H[k - 2], H[k - 1], P[i]) < 0) k--;
H[k++] = P[i];
}

// Remove duplicates
H.resize(k);
sort(H.begin(), H.end(), cmp);
H.erase(unique(H.begin(), H.end(), equ), H.end());
return H;
}
};
``````

Python version:

``````# http://www.algorithmist.com/index.php/Monotone_Chain_Convex_Hull.py

class Solution(object):

def outerTrees(self, points):
"""Computes the convex hull of a set of 2D points.

Input: an iterable sequence of (x, y) pairs representing the points.
Output: a list of vertices of the convex hull in counter-clockwise order,
starting from the vertex with the lexicographically smallest coordinates.
Implements Andrew's monotone chain algorithm. O(n log n) complexity.
"""

# Sort the points lexicographically (tuples are compared lexicographically).
# Remove duplicates to detect the case we have just one unique point.
# points = sorted(set(points))
points = sorted(points, key=lambda p: (p.x, p.y))

# Boring case: no points or a single point, possibly repeated multiple times.
if len(points) <= 1:
return points

# 2D cross product of OA and OB vectors, i.e. z-component of their 3D cross product.
# Returns a positive value, if OAB makes a counter-clockwise turn,
# negative for clockwise turn, and zero if the points are collinear.
def cross(o, a, b):
# return (a[0] - o[0]) * (b[1] - o[1]) - (a[1] - o[1]) * (b[0] - o[0])
return (a.x - o.x) * (b.y - o.y) - (a.y - o.y) * (b.x - o.x)

# Build lower hull
lower = []
for p in points:
while len(lower) >= 2 and cross(lower[-2], lower[-1], p) < 0:
lower.pop()
lower.append(p)

# Build upper hull
upper = []
for p in reversed(points):
while len(upper) >= 2 and cross(upper[-2], upper[-1], p) < 0:
upper.pop()
upper.append(p)

# Concatenation of the lower and upper hulls gives the convex hull.
# Last point of each list is omitted because it is repeated at the
# beginning of the other list.
# return lower[:-1] + upper[:-1]
return list(set(lower[:-1] + upper[:-1]))
``````

• This solution looks more concise, I do not know why it did not get any votes or comments.

• Great solution using Monotone Chain! Time complexity O(nlogn). I prefer this method to Jarvis Algorithm, which is O(n^2), and Graham scan, which is more difficult to prove and to implement. Thanks!

Here is my C++ code with a little bit optimization. After lower hull, you can check whether answer.size() == n for the single line case/duplicates.

``````class Solution {
public:
vector<Point> outerTrees(vector<Point>& points) {
// Andrew's monotone chain method
int n = points.size();
vector<Point> ans;
sort(points.begin(), points.end(), mycompare);
// left to right
for (int i = 0; i < n; ++i) {
while (ans.size() > 1 && orientation(ans[ans.size()-2], ans.back(), points[i]) < 0)
ans.pop_back();
ans.push_back(points[i]);
}
// if all points along a line, ans.size() is n after left to right procedure
if (ans.size() == n) return ans;
// right to left
for (int i = n-2; i >= 0; --i) {
while (ans.size() > 1 && orientation(ans[ans.size()-2], ans.back(), points[i]) < 0)
ans.pop_back();
ans.push_back(points[i]);
}
ans.pop_back();
return ans;
}
private:
static bool mycompare(Point& a, Point& b) {
return a.x < b.x || (a.x == b.x && a.y < b.y);
}
int orientation(Point& a, Point& b, Point& c) {
return (b.x-a.x)*(c.y-b.y) - (b.y-a.y)*(c.x-b.x);
}
};
``````

• @zestypanda wow, impressive, your solution is even more self-explained.

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