LeetCode #3454 — HARD

Separate Squares II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

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The Problem

Problem Statement

You are given a 2D integer array squares. Each squares[i] = [xi, yi, li] 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 covered by squares above the line equals the total area covered by squares below the line.

Answers within 10-5 of the actual answer will be accepted.

Note: Squares may overlap. Overlapping areas should be counted only once in this version.

Example 1:

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

Output: 1.00000

Explanation:

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

Explanation:

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.
Patterns Used
🔍Modified Binary Search→📐Segment Tree→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given a 2D integer array squares. Each squares[i] = [xi, yi, li] 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 covered by squares above the line equals the total area covered by squares below the line. Answers within 10-5 of the actual answer will be accepted. Note: Squares may overlap. Overlapping areas should be counted only once in this version.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Segment Tree

Example 1

[[0,0,1],[2,2,1]]

Example 2

[[0,0,2],[1,1,1]]

Related Problems

  • Rectangle Area II (rectangle-area-ii)
Step 02

Core Insight

What unlocks the optimal approach

  • Use a line sweep and a segment tree.
  • The line must lie in one of the squares.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3454: Separate Squares II
class Node {
    int l, r, cnt, length;
}

class SegmentTree {
    private Node[] tr;
    private int[] nums;

    public SegmentTree(int[] nums) {
        this.nums = nums;
        int n = nums.length - 1;
        tr = new Node[n << 2];
        for (int i = 0; i < tr.length; ++i) {
            tr[i] = new Node();
        }
        build(1, 0, n - 1);
    }

    private void build(int u, int l, int r) {
        tr[u].l = l;
        tr[u].r = r;
        if (l != r) {
            int mid = (l + r) >> 1;
            build(u << 1, l, mid);
            build(u << 1 | 1, mid + 1, r);
        }
    }

    public void modify(int u, int l, int r, int k) {
        if (tr[u].l >= l && tr[u].r <= r) {
            tr[u].cnt += k;
        } else {
            int mid = (tr[u].l + tr[u].r) >> 1;
            if (l <= mid) {
                modify(u << 1, l, r, k);
            }
            if (r > mid) {
                modify(u << 1 | 1, l, r, k);
            }
        }
        pushup(u);
    }

    private void pushup(int u) {
        if (tr[u].cnt > 0) {
            tr[u].length = nums[tr[u].r + 1] - nums[tr[u].l];
        } else if (tr[u].l == tr[u].r) {
            tr[u].length = 0;
        } else {
            tr[u].length = tr[u << 1].length + tr[u << 1 | 1].length;
        }
    }

    public int query() {
        return tr[1].length;
    }
}

class Solution {
    public double separateSquares(int[][] squares) {
        Set<Integer> xs = new HashSet<>();
        List<int[]> segs = new ArrayList<>();
        for (int[] sq : squares) {
            int x1 = sq[0], y1 = sq[1], l = sq[2];
            int x2 = x1 + l, y2 = y1 + l;
            xs.add(x1);
            xs.add(x2);
            segs.add(new int[] {y1, x1, x2, 1});
            segs.add(new int[] {y2, x1, x2, -1});
        }
        segs.sort(Comparator.comparingInt(a -> a[0]));
        int[] st = new int[xs.size()];
        int i = 0;
        for (int x : xs) {
            st[i++] = x;
        }
        Arrays.sort(st);
        SegmentTree tree = new SegmentTree(st);
        Map<Integer, Integer> d = new HashMap<>(st.length);
        for (i = 0; i < st.length; i++) {
            d.put(st[i], i);
        }
        double area = 0.0;
        int y0 = 0;
        for (int[] s : segs) {
            int y = s[0], x1 = s[1], x2 = s[2], k = s[3];
            area += (double) (y - y0) * tree.query();
            tree.modify(1, d.get(x1), d.get(x2) - 1, k);
            y0 = y;
        }
        double target = area / 2.0;
        area = 0.0;
        y0 = 0;
        for (int[] s : segs) {
            int y = s[0], x1 = s[1], x2 = s[2], k = s[3];
            double t = (double) (y - y0) * tree.query();
            if (area + t >= target) {
                return y0 + (target - area) / tree.query();
            }
            area += t;
            tree.modify(1, d.get(x1), d.get(x2) - 1, k);
            y0 = y;
        }
        return 0.0;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n × log n)
Space
O(n)

Approach Breakdown

LINEAR SCAN
O(n) time
O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH
O(log n) time
O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Related Problems
#315Count of Smaller Numbers After Selfhard→#327Count of Range Sumhard→#363Max Sum of Rectangle No Larger Than Khard→#456132 Patternmedium→#493Reverse Pairshard→