Given a set of points, the task is to find the covex hull of the given points. The convex hull is the smallest convex polygon that contains all the points.

Please check this article first: Convex Hull | Set 1 (Jarvis’s Algorithm or Wrapping)

**Examples:**

Input:Points[] = {{0, 3}, {2, 2}, {1, 1}, {2, 1}, {3, 0}, {0, 0}, {3, 3}}

Output:

(0, 0)

(3, 0)

(3, 3)

(0, 3)

**Approach:** Monotone chain algorithm constructs the convex hull in **O(n * log(n))** time. We have to sort the points first and then calculate the upper and lower hulls in **O(n)** time. The points will be sorted with respect to x-coordinates (with respect to y-coordinates in case of a tie in x-coordinates), we will then find the left most point and then try to rotate in clockwise direction and find the next point and then repeat the step until we reach the rightmost point and then again rotate in the the clockwise direction and find the lower hull.

Below is the implementation of the above approach:

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `#define llu long long int ` `using` `namespace` `std; ` ` ` `struct` `Point { ` ` ` ` ` `llu x, y; ` ` ` ` ` `bool` `operator<(Point p) ` ` ` `{ ` ` ` `return` `x < p.x || (x == p.x && y < p.y); ` ` ` `} ` `}; ` ` ` `// Cross product of two vectors OA and OB ` `// returns positive for counter clockwise ` `// turn and negative for clockwise turn ` `llu cross_product(Point O, Point A, Point B) ` `{ ` ` ` `return` `(A.x - O.x) * (B.y - O.y) ` ` ` `- (A.y - O.y) * (B.x - O.x); ` `} ` ` ` `// Returns a list of points on the convex hull ` `// in counter-clockwise order ` `vector<Point> convex_hull(vector<Point> A) ` `{ ` ` ` `int` `n = A.size(), k = 0; ` ` ` ` ` `if` `(n <= 3) ` ` ` `return` `A; ` ` ` ` ` `vector<Point> ans(2 * n); ` ` ` ` ` `// Sort points lexicographically ` ` ` `sort(A.begin(), A.end()); ` ` ` ` ` `// Build lower hull ` ` ` `for` `(` `int` `i = 0; i < n; ++i) { ` ` ` ` ` `// If the point at K-1 position is not a part ` ` ` `// of hull as vector from ans[k-2] to ans[k-1] ` ` ` `// and ans[k-2] to A[i] has a clockwise turn ` ` ` `while` `(k >= 2 && cross_product(ans[k - 2], ` ` ` `ans[k - 1], A[i]) <= 0) ` ` ` `k--; ` ` ` `ans[k++] = A[i]; ` ` ` `} ` ` ` ` ` `// Build upper hull ` ` ` `for` `(` `size_t` `i = n - 1, t = k + 1; i > 0; --i) { ` ` ` ` ` `// If the point at K-1 position is not a part ` ` ` `// of hull as vector from ans[k-2] to ans[k-1] ` ` ` `// and ans[k-2] to A[i] has a clockwise turn ` ` ` `while` `(k >= t && cross_product(ans[k - 2], ` ` ` `ans[k - 1], A[i - 1]) <= 0) ` ` ` `k--; ` ` ` `ans[k++] = A[i - 1]; ` ` ` `} ` ` ` ` ` `// Resize the array to desired size ` ` ` `ans.resize(k - 1); ` ` ` ` ` `return` `ans; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `vector<Point> points; ` ` ` ` ` `// Add points ` ` ` `points.push_back({ 0, 3 }); ` ` ` `points.push_back({ 2, 2 }); ` ` ` `points.push_back({ 1, 1 }); ` ` ` `points.push_back({ 2, 1 }); ` ` ` `points.push_back({ 3, 0 }); ` ` ` `points.push_back({ 0, 0 }); ` ` ` `points.push_back({ 3, 3 }); ` ` ` ` ` `// Find the convex hull ` ` ` `vector<Point> ans = convex_hull(points); ` ` ` ` ` `// Print the convex hull ` ` ` `for` `(` `int` `i = 0; i < ans.size(); i++) ` ` ` `cout << ` `"("` `<< ans[i].x << ` `", "` ` ` `<< ans[i].y << ` `")"` `<< endl; ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

(0, 0) (3, 0) (3, 3) (0, 3)

## Recommended Posts:

- Dynamic Convex hull | Adding Points to an Existing Convex Hull
- Convex Hull | Set 1 (Jarvis's Algorithm or Wrapping)
- Quickhull Algorithm for Convex Hull
- Convex Hull using Divide and Conquer Algorithm
- Convex Hull | Set 2 (Graham Scan)
- Deleting points from Convex Hull
- Perimeter of Convex hull for a given set of points
- Tangents between two Convex Polygons
- Find number of diagonals in n sided convex polygon
- Check whether two convex regular polygon have same center or not
- Check if the given point lies inside given N points of a Convex Polygon
- Check if given polygon is a convex polygon or not
- Chain Code for 2D Line
- Maximum Length Chain of Pairs | Set-2
- Algorithm Library | C++ Magicians STL Algorithm
- Which sorting algorithm makes minimum number of memory writes?
- Closest Pair of Points using Divide and Conquer algorithm
- Know Your Sorting Algorithm | Set 1 (Sorting Weapons used by Programming Languages)
- Know Your Sorting Algorithm | Set 2 (Introsort- C++’s Sorting Weapon)
- Klee's Algorithm (Length Of Union Of Segments of a line)

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