#3453

Medium

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Separate Squares I

Find the minimum y-coordinate of a horizontal line that balances the areas of squares above and below it.

Problem Statement

You are given a 2D array, squares, where each element squares[i] = [xi, yi, li] represents a square with the bottom-left corner at (xi, yi) and a side length of li. Your goal is to find the minimum y-coordinate of a horizontal line such that the total area of the squares above the line equals the total area of the squares below it.

The problem allows answers within a tolerance of 10^-5 from the actual result. This requires careful handling of floating-point arithmetic and the efficient use of binary search on the valid answer space to minimize the difference in areas above and below the horizontal line.

Examples

Example 1

Input: squares = [[0,0,1],[2,2,1]]

Output: 1.00000

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.

Example 2

Input: squares = [[0,0,2],[1,1,1]]

Output: 1.16667

The areas are: Since the areas above and below the line are equal, the output is 7/6 = 1.16667 .

Constraints

1 <= squares.length <= 5 * 104
squares[i] = [xi, yi, li]
squares[i].length == 3
0 <= xi, yi <= 109
1 <= li <= 109
The total area of all the squares will not exceed 1012.

Solution Approach

Binary Search Over Answer Space

Apply binary search to the range of possible y-coordinates to find the minimum y that balances the areas. For each candidate y, calculate the area of the squares above and below the line and adjust the search range based on the comparison.

Calculate Areas Efficiently

For each candidate y, calculate the areas of the squares that are above and below it. This requires considering each square's y-coordinate and side length to determine whether it contributes to the area above or below the line.

Precision Handling

Since the answer must be accurate to 10^-5, use a precise floating-point comparison to determine when the areas above and below the line are sufficiently balanced. Ensure the solution can handle large inputs efficiently.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the binary search over the y-coordinate range and the area calculations. The binary search will require log(n) iterations, where n is the size of the range of possible y-values, and in each iteration, the areas must be recalculated for all squares, leading to a complexity of O(n log M), where M is the size of the y-range and n is the number of squares.

What Interviewers Usually Probe

Assess understanding of binary search over valid answer space.
Check how the candidate handles floating-point precision in area calculations.
Evaluate the candidate's approach to optimizing performance for large inputs.

Common Pitfalls or Variants

Common pitfalls

Failing to correctly handle floating-point precision when comparing areas.
Overcomplicating the area calculations instead of using a simple and efficient method.
Not adjusting the search range in binary search properly, leading to incorrect results.

Follow-up variants

Handling larger numbers of squares efficiently.
Modifying the problem to account for non-axis-aligned squares.
Improving performance with specialized techniques for faster area computation.

FAQ

How does binary search apply to Separate Squares I?

Binary search is used to find the minimum y-coordinate where the areas of the squares above and below the horizontal line are balanced.

What is the time complexity of solving this problem?

The time complexity is O(n log M), where n is the number of squares and M is the size of the range of y-values.

How do you calculate the area above and below the line?

For each square, check if its top part lies above or below the line and calculate the area accordingly.

What precision is required for the answer?

The answer must be correct within 10^-5 of the actual value.

What are the common mistakes when solving Separate Squares I?

Common mistakes include incorrect floating-point comparisons, inefficient area calculations, and improper binary search adjustments.

terminal

Solution

Solution 1: Binary Search

According to the problem, we need to find a horizontal line such that the total area of squares above the line equals the total area of squares below the line. Since as the $y$ coordinate increases, the area below the line increases and the area above the line decreases, we can use binary search to find the $y$ coordinate of this horizontal line.

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class Solution:
    def separateSquares(self, squares: List[List[int]]) -> float:
        def check(y1: float) -> bool:
            t = 0
            for _, y, l in squares:
                if y < y1:
                    t += l * min(y1 - y, l)
            return t >= s / 2

        s = sum(a[2] * a[2] for a in squares)
        l, r = 0, max(a[1] + a[2] for a in squares)
        eps = 1e-5
        while r - l > eps:
            mid = (l + r) / 2
            if check(mid):
                r = mid
            else:
                l = mid
        return r

Continue Practicing

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