Separate Squares I

Naive Solution

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.

Optimal Solution

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.

1. Determine the Search Range: Find the minimum and maximum y values among all squares. The search will be performed within this range.
2. Binary Search: Iteratively narrow the search range by halving it in each step.
3. Calculate Areas: For a given y (mid-point in the search range), iterate through the squares and compute the area above and below the line.
4. Adjust Search Range: If the area above the line is greater than the area below, the target y is lower. Otherwise, the target y is higher.
5. Termination: Stop the search when the difference between above and below areas is within the tolerance (10^-5).

Edge Cases:

• Empty input: Handle the case where the input array squares is empty. Return 0 in this case.
• Single square: If there is only one square, the median will be at y + l/2, return the bottom left y coordinate + half of the length
• Overlapping squares: The algorithm should correctly account for overlapping squares. The areas are additive.

Code 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.