You are given an integer n
representing the dimensions of an n x n
grid, with the origin at the bottom-left corner of the grid. You are also given a 2D array of coordinates rectangles
, where rectangles[i]
is in the form [startx, starty, endx, endy]
, representing a rectangle on the grid. Each rectangle is defined as follows:
(startx, starty)
: The bottom-left corner of the rectangle.(endx, endy)
: The top-right corner of the rectangle.Note that the rectangles do not overlap. Your task is to determine if it is possible to make either two horizontal or two vertical cuts on the grid such that:
Return true
if such cuts can be made; otherwise, return false
.
For example:
n = 5, rectangles = [[1,0,5,2],[0,2,2,4],[3,2,5,3],[0,4,4,5]]
should return true
because we can make horizontal cuts at y = 2
and y = 4
.
n = 4, rectangles = [[0,0,1,1],[2,0,3,4],[0,2,2,3],[3,0,4,3]]
should return true
because we can make vertical cuts at x = 2
and x = 3
.
n = 4, rectangles = [[0,2,2,4],[1,0,3,2],[2,2,3,4],[3,0,4,2],[3,2,4,4]]
should return false
because we cannot make two horizontal or two vertical cuts that satisfy the conditions.
What approach would you take to solve this problem? What is the time and space complexity of your solution?
A brute force approach would involve iterating through all possible pairs of horizontal and vertical cuts and checking if the resulting sections satisfy the given conditions. We can generate all possible pairs of cuts and then iterate through the rectangles to check which section they belong to. This would have a high time complexity.
We can optimize the solution by first trying horizontal cuts and then vertical cuts. For each direction, we iterate through all possible pairs of cuts. For a given pair of cuts, we check if each of the three sections contains at least one rectangle and that every rectangle belongs to exactly one section.
y
values for the first cut (from 1 to n-2).y
values for the second cut (from first cut + 1 to n-1).x
values for the first cut (from 1 to n-2).x
values for the second cut (from first cut + 1 to n-1).Here's a Python implementation of the optimal solution:
def solve():
n = 5
rectangles = [[1, 0, 5, 2], [0, 2, 2, 4], [3, 2, 5, 3], [0, 4, 4, 5]]
def check_cuts(cuts, horizontal):
sections = [[] for _ in range(3)]
for rect in rectangles:
start_x, start_y, end_x, end_y = rect
if horizontal:
if start_y < cuts[0]:
sections[0].append(rect)
elif start_y < cuts[1]:
sections[1].append(rect)
else:
sections[2].append(rect)
else:
if start_x < cuts[0]:
sections[0].append(rect)
elif start_x < cuts[1]:
sections[1].append(rect)
else:
sections[2].append(rect)
for section in sections:
if not section:
return False
all_rects = []
for section in sections:
all_rects.extend(section)
if len(all_rects) != len(rectangles):
return False
return True
def check_horizontal_cuts():
for i in range(1, n - 1):
for j in range(i + 1, n):
if check_cuts([i, j], True):
return True
return False
def check_vertical_cuts():
for i in range(1, n - 1):
for j in range(i + 1, n):
if check_cuts([i, j], False):
return True
return False
return check_horizontal_cuts() or check_vertical_cuts()
print(solve())
n^2
comes from iterating through pairs of cuts and m
is iterating through the rectangles to check if they satisfy the condition. In the worst case, the loops go through all possible n
values, and the check cuts function goes through each rectangle, therefore the time complexity is O(n^2 * m).check_cuts
function, the space complexity is O(m) because in the worst case, all rectangles will be added to these lists.start_x >= end_x
or start_y >= end_y
), the algorithm should handle them gracefully (e.g., by returning false
or raising an exception).