LeetCode #3453 — MEDIUM

Separate Squares I

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given a 2D integer array squares. Each squares[i] = [x_i, y_i, l_i] 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^-5 of the actual answer will be accepted.

Note: Squares may overlap. Overlapping areas should be counted multiple times.

Example 1:

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

Output: 1.00000

Explanation:

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

Explanation:

The areas are:

Below the line: 7/6 * 2 (Red) + 1/6 (Blue) = 15/6 = 2.5.
Above the line: 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.

Constraints:

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

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 of the squares above the line equals the total area of the squares below the line. Answers within 10-5 of the actual answer will be accepted. Note: Squares may overlap. Overlapping areas should be counted multiple times.

Baseline thinking

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

Pattern signal: Array · Binary Search

Example 1

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

Example 2

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

Step 02

Core Insight

What unlocks the optimal approach

Binary search on the answer.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
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

Upper-end input sizes

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 #3453: Separate Squares I
class Solution {
    private int[][] squares;
    private double s;

    private boolean check(double y1) {
        double t = 0.0;
        for (int[] a : squares) {
            int y = a[1];
            int l = a[2];
            if (y < y1) {
                t += (double) l * Math.min(y1 - y, l);
            }
        }
        return t >= s / 2.0;
    }

    public double separateSquares(int[][] squares) {
        this.squares = squares;
        s = 0.0;
        double l = 0.0;
        double r = 0.0;
        for (int[] a : squares) {
            s += (double) a[2] * a[2];
            r = Math.max(r, a[1] + a[2]);
        }

        double eps = 1e-5;
        while (r - l > eps) {
            double mid = (l + r) / 2.0;
            if (check(mid)) {
                r = mid;
            } else {
                l = mid;
            }
        }
        return r;
    }
}

// Accepted solution for LeetCode #3453: Separate Squares I
func separateSquares(squares [][]int) float64 {
	s := 0.0
	check := func(y1 float64) bool {
		t := 0.0
		for _, a := range squares {
			y := a[1]
			l := a[2]
			if float64(y) < y1 {
				h := min(float64(l), y1-float64(y))
				t += float64(l) * h
			}
		}
		return t >= s/2.0
	}
	l, r := 0.0, 0.0
	for _, a := range squares {
		s += float64(a[2] * a[2])
		r = max(r, float64(a[1]+a[2]))
	}

	const eps = 1e-5
	for r-l > eps {
		mid := (l + r) / 2.0
		if check(mid) {
			r = mid
		} else {
			l = mid
		}
	}
	return r
}

# Accepted solution for LeetCode #3453: Separate Squares I
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

// Accepted solution for LeetCode #3453: Separate Squares I
impl Solution {
    pub fn separate_squares(squares: Vec<Vec<i32>>) -> f64 {
        let mut s: f64 = 0.0;

        let mut l: f64 = 0.0;
        let mut r: f64 = 0.0;

        for a in squares.iter() {
            let len = a[2] as f64;
            s += len * len;
            r = r.max((a[1] + a[2]) as f64);
        }

        let check = |y1: f64| -> bool {
            let mut t: f64 = 0.0;
            for a in squares.iter() {
                let y = a[1] as f64;
                let l = a[2] as f64;
                if y < y1 {
                    let h = l.min(y1 - y);
                    t += l * h;
                }
            }
            t >= s / 2.0
        };

        const EPS: f64 = 1e-5;
        while r - l > EPS {
            let mid = (l + r) / 2.0;
            if check(mid) {
                r = mid;
            } else {
                l = mid;
            }
        }
        r
    }
}

// Accepted solution for LeetCode #3453: Separate Squares I
function separateSquares(squares: number[][]): number {
    const check = (y1: number): boolean => {
        let t = 0;
        for (const [_, y, l] of squares) {
            if (y < y1) {
                t += l * Math.min(y1 - y, l);
            }
        }
        return t >= s / 2;
    };

    let s = 0;
    let l = 0;
    let r = 0;
    for (const a of squares) {
        s += a[2] * a[2];
        r = Math.max(r, a[1] + a[2]);
    }

    const eps = 1e-5;
    while (r - l > eps) {
        const mid = (l + r) / 2;
        if (check(mid)) {
            r = mid;
        } else {
            l = mid;
        }
    }
    return r;
}

Step 06

Interactive Study Demo

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

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

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.