You are given a 2D integer array squares
. Each squares[i] = [x<sub>i</sub>, y<sub>i</sub>, l<sub>i</sub>]
represents the coordinates of the bottom-left point and the side length of a square parallel to the x-axis.
Find the minimum y-coordinate value of a horizontal line such that the total area of the squares above the line equals the total area of the squares below the line.
Answers within 10<sup>-5</sup>
of the actual answer will be accepted.
Note: Squares may overlap. Overlapping areas should be counted multiple times.
For example:
squares = [[0,0,1],[2,2,1]]
In this example, any horizontal line between y = 1
and y = 2
will have 1 square unit above it and 1 square unit below it. The lowest option is 1.
Here is another example:
squares = [[0,0,2],[1,1,1]]
The areas are:
7/6 * 2 (Red) + 1/6 (Blue) = 15/6 = 2.5
.5/6 * 2 (Red) + 5/6 (Blue) = 15/6 = 2.5
.Since the areas above and below the line are equal, the output is 7/6 = 1.16667
.
Write an efficient function to solve this problem. Describe the time and space complexity of your solution. Also, consider edge cases.
A brute-force approach would involve iterating through a large number of possible y-coordinate values and, for each value, calculating the total area of squares above and below the line. The y-coordinate that makes the areas equal is the answer. However, this is highly inefficient due to the potentially infinite number of y-values and the need to recompute areas for each y.
Time Complexity: O(N * K), where N is the number of squares and K is the number of iterations (y-values tested). K can be extremely large if high precision is needed.
Space Complexity: O(1), as we only need a few variables to store areas and the current y-coordinate.
A more efficient approach involves using binary search. The range for the binary search is determined by the minimum and maximum y-coordinates of the squares. Within the binary search, we calculate the total area above and below the mid-point y
. Based on whether the above
area is greater or less than the below
area, we adjust our search range accordingly.
y
values among all squares. The search will be performed within this range.y
(mid-point in the search range), iterate through the squares and compute the area above and below the line.y
is lower. Otherwise, the target y
is higher.above
and below
areas is within the tolerance (10^-5
).Edge Cases:
squares
is empty. Return 0 in this case.y + l/2
, return the bottom left y
coordinate + half of the lengthCode Example (Python):
def find_minimum_y(squares):
if not squares:
return 0.0
min_y = min(y for _, y, _ in squares)
max_y = max(y + l for _, y, l in squares)
tolerance = 1e-5
while max_y - min_y > tolerance:
mid_y = (min_y + max_y) / 2.0
area_above = 0.0
area_below = 0.0
for x, y, l in squares:
if mid_y <= y:
area_above += l * l
elif mid_y >= y + l:
area_below += l * l
else:
above_height = y + l - mid_y
below_height = mid_y - y
area_above += above_height * l
area_below += below_height * l
if area_above > area_below:
min_y = mid_y
else:
max_y = mid_y
return min_y
Time Complexity: O(N * log(R)), where N is the number of squares and R is the range of y-values. The binary search performs log(R) iterations, and in each iteration, we iterate through all the squares (O(N)).
Space Complexity: O(1), as we only use a constant amount of extra space to store variables.