587. Erect the Fence

# Time:  O(nlogn)
# Space: O(n)

# Monotone Chain Algorithm
# Template: https://en.wikibooks.org/wiki/Algorithm_Implementation/Geometry/Convex_hull/Monotone_chain#Python
class Solution(object):
    def outerTrees(self, points):
        """
        :type points: List[List[int]]
        :rtype: List[List[int]]
        """
        # Sort the points lexicographically (tuples are compared lexicographically).
        # Remove duplicates to detect the case we have just one unique point.
        points = sorted(set(tuple(x) for x in points))

        # 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])

        # Build lower hull 
        lower = []
        for p in points:
            while len(lower) >= 2 and cross(lower[-2], lower[-1], p) < 0:  # modified
                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:  # modified
                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. 
        result = lower[:-1] + upper[:-1]
        return result if result[1] != result[-1] else result[:len(result)//2+1]  # modified

Solution from kamyu104/LeetCode-Solutions · MIT