You are given an array trees where trees[i] = [xi, yi] represents the location of a tree in the garden.
Fence the entire garden using the minimum length of rope, as it is expensive. The garden is well-fenced only if all the trees are enclosed.
Return the coordinates of trees that are exactly located on the fence perimeter. You may return the answer in any order.
Example 1:
Input: trees = [[1,1],[2,2],[2,0],[2,4],[3,3],[4,2]] Output: [[1,1],[2,0],[4,2],[3,3],[2,4]] Explanation: All the trees will be on the perimeter of the fence except the tree at [2, 2], which will be inside the fence.
Example 2:
Input: trees = [[1,2],[2,2],[4,2]] Output: [[4,2],[2,2],[1,2]] Explanation: The fence forms a line that passes through all the trees.
Constraints:
1 <= trees.length <= 3000trees[i].length == 20 <= xi, yi <= 100When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
Imagine you have many points scattered on a field, and you want to build a fence that encloses all of them. The brute force method involves trying every possible combination of points to see if they form a fence that meets the requirements. We systematically check each potential fence until we find one that works.
Here's how the algorithm would work step-by-step:
from itertools import combinations
def erect_the_fence_brute_force(points):
potential_fences = []
number_of_points = len(points)
# Iterate through possible fence sizes
for fence_size in range(3, number_of_points + 1):
# Generate all combinations of points
for point_indices in combinations(range(number_of_points), fence_size):
fence_points = [points[index] for index in point_indices]
remaining_points = [points[i] for i in range(number_of_points) if i not in point_indices]
# Check if all remaining points are inside the fence
if is_valid_fence(fence_points, remaining_points):
potential_fences.append(fence_points)
# Find the fence with the smallest area (or perimeter, etc.)
if not potential_fences:
return []
best_fence = min(potential_fences, key=calculate_area)
return best_fence
def is_valid_fence(fence_points, remaining_points):
if len(fence_points) == 3:
#For triangles, check all points are inside
for point in remaining_points:
if not is_point_inside_triangle(point, fence_points[0], fence_points[1], fence_points[2]):
return False
return True
#For more complex shapes, check all points are inside
convex_hull_points = convex_hull(fence_points)
for point in remaining_points:
if not is_point_inside_polygon(point, convex_hull_points):
return False
return True
def convex_hull(points):
points = sorted(set(points))
if len(points) <= 1:
return points
def cross(o, a, b):
return (a[0] - o[0]) * (b[1] - o[1]) - (a[1] - o[1]) * (b[0] - o[0])
lower = []
for point in points:
while len(lower) >= 2 and cross(lower[-2], lower[-1], point) <= 0:
lower.pop()
lower.append(point)
upper = []
for point in reversed(points):
while len(upper) >= 2 and cross(upper[-2], upper[-1], point) <= 0:
upper.pop()
upper.append(point)
return lower[:-1] + upper[:-1]
def is_point_inside_triangle(point, a, b, c):
def sign(p1, p2, p3):
return (p1[0] - p3[0]) * (p2[1] - p3[1]) - (p2[0] - p3[0]) * (p1[1] - p3[1])
d1 = sign(point, a, b)
d2 = sign(point, b, c)
d3 = sign(point, c, a)
has_neg = (d1 < 0) or (d2 < 0) or (d3 < 0)
has_pos = (d1 > 0) or (d2 > 0) or (d3 > 0)
return not (has_neg and has_pos)
def is_point_inside_polygon(point, polygon_points):
number_of_points = len(polygon_points)
inside = False
p1x, p1y = polygon_points[0]
for i in range(1, number_of_points + 1):
p2x, p2y = polygon_points[i % number_of_points]
if point[1] > min(p1y, p2y):
if point[1] <= max(p1y, p2y):
if point[0] <= max(p1x, p2x):
if p1y != p2y:
intersection_x = (point[1] - p1y) * (p2x - p1x) / (p2y - p1y) + p1x
if p1x == p2x or point[0] <= intersection_x:
inside = not inside
p1x, p1y = p2x, p2y
return inside
def calculate_area(fence_points):
#Dummy implementation. Calculate real area here.
return len(fence_points)
# Example usage (replace with your actual points)
# points = [[1, 1], [2, 2], [2, 0], [2, 4], [3, 3], [4, 2]]
# fence = erect_the_fence_brute_force(points)
# print(fence)The most efficient way to solve this problem is to imagine wrapping a rubber band around all the trees. The rubber band represents the fence, and we need to find which trees the rubber band will touch. This is achieved by identifying the trees that form the outer boundary of the tree arrangement.
Here's how the algorithm would work step-by-step:
def erect_the_fence(trees):
def cross_product_length(a, b, c):
return (b[0] - a[0]) * (c[1] - a[1]) - (b[1] - a[1]) * (c[0] - a[0])
if len(trees) < 3:
return trees
# Sort points to easily construct the convex hull
trees.sort(key=lambda point: (point[0], point[1]))
lower_hull = []
for point in trees:
# Maintain counter-clockwise order in the lower hull
while len(lower_hull) >= 2 and cross_product_length(lower_hull[-2], lower_hull[-1], point) < 0:
lower_hull.pop()
lower_hull.append(point)
upper_hull = []
for point in reversed(trees):
# Maintain counter-clockwise order in the upper hull
while len(upper_hull) >= 2 and cross_product_length(upper_hull[-2], upper_hull[-1], point) < 0:
upper_hull.pop()
upper_hull.append(point)
# Combine the hulls, removing duplicates
convex_hull = lower_hull[:-1] + upper_hull[:-1]
# Remove duplicate points from the result.
return list(set(map(tuple, convex_hull)))| Case | How to Handle |
|---|---|
| Empty input list of points | Return an empty list immediately as there are no points to form a fence. |
| Input list contains only one or two points | Return the original input list as a fence can be formed with these points. |
| All points are collinear (lie on the same line) | Graham scan handles this by selecting the extreme points and adding intermediate collinear points on the hull boundary, ensuring all points are included in the fence. |
| Duplicate points in the input list | The Graham scan algorithm considers all points, and duplicates will simply be part of the convex hull. |
| Points forming a rectangle or other simple polygon | Graham scan correctly identifies the vertices of the polygon as the fence. |
| Large number of input points (performance considerations) | Graham scan has a time complexity of O(n log n) due to the sorting step, which should be efficient enough for reasonably sized inputs. |
| Integer overflow during cross-product calculation | Use 64-bit integers (long) for cross-product calculations to prevent overflow, especially when dealing with large coordinate values. |
| Extreme coordinate values (very large or very small) | Ensure the chosen data types (e.g., long) can accommodate the range of coordinate values to prevent overflow or loss of precision. |