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Erect the Fence

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ArraysGreedy Algorithms

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 <= 3000
  • trees[i].length == 2
  • 0 <= xi, yi <= 100
  • All the given positions are unique.

Solution


Clarifying Questions

When 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:

  1. What data type will the point coordinates be (integers or floats)? What's the range of possible coordinate values?
  2. If all points are collinear (i.e., form a straight line), what should the output be? Should I return all of them, or is it an invalid input?
  3. Are duplicate points allowed in the input? If so, how should they be handled in the output – should the fence include all duplicates, or just one of each?
  4. If multiple valid fences with the minimum perimeter exist, is any one of them acceptable?
  5. Is the order of points in the returned fence significant? If so, what ordering criteria should I use (e.g., clockwise, counter-clockwise, lexicographical order)?

Brute Force Solution

Approach

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:

  1. Pick any three points from the group of points.
  2. Check if all the remaining points are inside the triangle formed by those three points.
  3. If all points are inside (or on the edges), you've found a potential fence (a bad one, but a fence).
  4. Repeat the above steps for every possible combination of three points.
  5. Now, try every possible combination of four points, checking if all other points are inside the shape created by these four points.
  6. Continue doing this for five points, six points, and so on, until you've considered all possible combinations of points forming the fence.
  7. From all the potential fences found, determine which one has the smallest area, or the shortest perimeter, or meets some other criteria for what makes a 'good' fence, and that is your final fence.

Code Implementation

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)

Big(O) Analysis

Time Complexity
O(n^n)The proposed algorithm iterates through all possible combinations of points to form potential fences. First, it considers all combinations of 3 points, then 4, and so on up to n points. For each k points, the number of combinations is n choose k, represented as n! / (k! * (n-k)!). The algorithm checks each combination to see if the remaining points are enclosed, requiring up to O(n) operations per combination. Summing the cost of all combinations from k=3 to k=n results in exponential time complexity, specifically O(n^n) in the worst case because it eventually must test all n points which has a polynomial runtime of at least O(n!) nested inside another loop with a polynomial runtime of O(n), resulting in an exponential function greater than O(n!).
Space Complexity
O(1)The provided brute force algorithm for constructing a fence primarily involves iterating through combinations of points and checking if the remaining points lie within the shape formed by those points. While it generates numerous potential fences, it doesn't explicitly store all of them simultaneously in a separate data structure. The process of checking points inside a shape and comparing areas would likely use a constant amount of extra space for variables to track the best fence found so far, area calculations, and point comparisons, independent of the number of input points N. Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

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:

  1. First, sort the trees by their location from left to right, and if some have the same left/right position, sort them by their up/down position.
  2. Next, build the lower part of the fence. Imagine walking from left to right, and keep track of the trees that would make a turn in the rubber band fence as you wrap around the bottom of the trees.
  3. Then, build the upper part of the fence. This is similar to the lower part, but you imagine walking from right to left and wrapping the rubber band around the top of the trees.
  4. Finally, combine the points from the lower and upper fences. Be careful to remove any duplicate trees that were included in both the upper and lower portions of the fence. This final set of trees is the minimal set needed to build your fence.

Code Implementation

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)))

Big(O) Analysis

Time Complexity
O(n log n)The dominant operation in this algorithm is the initial sorting of the trees, which takes O(n log n) time. Building the lower and upper hulls involves iterating through the sorted trees, which takes O(n) time each. While the orientation check inside the loop might appear to add complexity, it only performs a constant number of operations per tree. Therefore, the sorting step dictates the overall time complexity.
Space Complexity
O(N)The algorithm sorts the input list of trees, which might require O(N) auxiliary space depending on the sorting algorithm used. The upper and lower hulls, stored as lists, can each contain up to N points in the worst-case scenario, where N is the number of trees. The final result also requires a list to store the fence points, adding to the overall auxiliary space. Therefore, the space complexity is dominated by the hull storage and the result list, giving us O(N) space.

Edge Cases

Empty input list of points
How to Handle:
Return an empty list immediately as there are no points to form a fence.
Input list contains only one or two points
How to Handle:
Return the original input list as a fence can be formed with these points.
All points are collinear (lie on the same line)
How to Handle:
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
How to Handle:
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
How to Handle:
Graham scan correctly identifies the vertices of the polygon as the fence.
Large number of input points (performance considerations)
How to Handle:
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
How to Handle:
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)
How to Handle:
Ensure the chosen data types (e.g., long) can accommodate the range of coordinate values to prevent overflow or loss of precision.