#3454

Hard

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Separate Squares II

Separate Squares II requires finding the minimum y-coordinate such that squares' areas are split evenly above and below a horizontal line.

array binary search segment tree line sweep

Problem Statement

You are given a list of squares represented by their bottom-left coordinates and side lengths. Each square is aligned parallel to the x-axis. Your task is to find the minimum y-coordinate of a horizontal line that splits the total area covered by the squares into two equal parts: one above the line and the other below.

The total area of the squares above and below the line should be as close as possible. The solution must be precise, with answers within 10^-5 of the correct answer being accepted.

Examples

Example 1

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

Output: 1.00000

Any horizontal line between y = 1 and y = 2 results in an equal split, with 1 square unit above and 1 square unit below. The minimum y-value is 1.

Example 2

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

Output: 1.00000

Since the blue square overlaps with the red square, it will not be counted again. Thus, the line y = 1 splits the squares into two equal parts.

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 1015.

Solution Approach

Binary Search over the Answer Space

Binary search is used over the possible y-coordinate values, narrowing down the range where the area split is balanced. The challenge is to efficiently find the point where the areas above and below are equal.

Line Sweep with Segment Tree

A line sweep combined with a segment tree allows for efficient area calculation as the horizontal line moves. Segment trees help track the areas dynamically as squares are considered for inclusion above or below the line.

Handling Overlaps

Overlapping squares must be carefully handled since their area should only be counted once. This adds complexity when calculating the area above or below the line.

Complexity Analysis

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

The time complexity depends on the binary search iterations and the segment tree operations. For each binary search step, we perform a segment tree update/query operation, leading to a complexity of O(log N) per binary search iteration, where N is the number of squares.

What Interviewers Usually Probe

Can the candidate explain why binary search is effective in this problem?
Does the candidate know how to efficiently manage area updates when using segment trees?
Is the candidate aware of the potential pitfalls with overlapping squares?

Common Pitfalls or Variants

Common pitfalls

Not handling square overlaps correctly, resulting in inaccurate area calculations.
Incorrect binary search boundaries leading to suboptimal or incorrect solutions.
Failure to efficiently use the segment tree for area updates during the line sweep process.

Follow-up variants

What if the number of squares increases significantly? Can the solution scale to handle larger inputs?
What happens if we modify the problem to split the area into a ratio other than 50/50?
How would the solution change if we were to account for square rotation or other transformations?

FAQ

What is the core algorithm used in the Separate Squares II problem?

The core algorithm involves binary search over the valid answer space, combined with a line sweep and segment tree for efficient area calculations.

How does binary search help in solving Separate Squares II?

Binary search helps by narrowing down the y-coordinate value where the areas above and below the line are equal. It reduces the search space effectively.

Why is a segment tree used in this problem?

A segment tree is used to efficiently calculate and update the area covered by squares as the line sweeps across the y-axis.

What should be the accuracy of the solution in Separate Squares II?

The solution must be accurate to within 10^-5 of the correct y-coordinate for the area split.

How can overlapping squares affect the solution in Separate Squares II?

Overlapping squares must be handled carefully to ensure their area is not double-counted, which can affect the accuracy of the area split.

terminal

Solution

Solution 1: Sweep Line

This problem can be solved using the sweep line algorithm to calculate the total area of all squares.

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class Node:
    __slots__ = ("l", "r", "cnt", "length")

    def __init__(self):
        self.l = self.r = 0
        self.cnt = self.length = 0


class SegmentTree:
    def __init__(self, nums):
        n = len(nums) - 1
        self.nums = nums
        self.tr = [Node() for _ in range(n << 2)]
        self.build(1, 0, n - 1)

    def build(self, u, l, r):
        self.tr[u].l, self.tr[u].r = l, r
        if l != r:
            mid = (l + r) >> 1
            self.build(u << 1, l, mid)
            self.build(u << 1 | 1, mid + 1, r)

    def modify(self, u, l, r, k):
        if self.tr[u].l >= l and self.tr[u].r <= r:
            self.tr[u].cnt += k
        else:
            mid = (self.tr[u].l + self.tr[u].r) >> 1
            if l <= mid:
                self.modify(u << 1, l, r, k)
            if r > mid:
                self.modify(u << 1 | 1, l, r, k)
        self.pushup(u)

    def pushup(self, u):
        if self.tr[u].cnt:
            self.tr[u].length = self.nums[self.tr[u].r + 1] - self.nums[self.tr[u].l]
        elif self.tr[u].l == self.tr[u].r:
            self.tr[u].length = 0
        else:
            self.tr[u].length = self.tr[u << 1].length + self.tr[u << 1 | 1].length

    @property
    def length(self):
        return self.tr[1].length


class Solution:
    def separateSquares(self, squares: List[List[int]]) -> float:
        xs = set()
        segs = []
        for x1, y1, l in squares:
            x2, y2 = x1 + l, y1 + l
            xs.update([x1, x2])
            segs.append((y1, x1, x2, 1))
            segs.append((y2, x1, x2, -1))
        segs.sort()
        st = sorted(xs)
        tree = SegmentTree(st)
        d = {x: i for i, x in enumerate(st)}
        area = 0
        y0 = 0
        for y, x1, x2, k in segs:
            area += (y - y0) * tree.length
            tree.modify(1, d[x1], d[x2] - 1, k)
            y0 = y

        target = area / 2
        area = 0
        y0 = 0
        for y, x1, x2, k in segs:
            t = (y - y0) * tree.length
            if area + t >= target:
                return y0 + (target - area) / tree.length
            area += t
            tree.modify(1, d[x1], d[x2] - 1, k)
            y0 = y
        return 0

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