LeetCode #492 — EASY

Construct the Rectangle

Build confidence with an intuition-first walkthrough focused on math fundamentals.

The Problem

Problem Statement

A web developer needs to know how to design a web page's size. So, given a specific rectangular web page’s area, your job by now is to design a rectangular web page, whose length L and width W satisfy the following requirements:

The area of the rectangular web page you designed must equal to the given target area.
The width W should not be larger than the length L, which means L >= W.
The difference between length L and width W should be as small as possible.

Return an array [L, W] where L and W are the length and width of the web page you designed in sequence.

Example 1:

Input: area = 4
Output: [2,2]
Explanation: The target area is 4, and all the possible ways to construct it are [1,4], [2,2], [4,1]. 
But according to requirement 2, [1,4] is illegal; according to requirement 3,  [4,1] is not optimal compared to [2,2]. So the length L is 2, and the width W is 2.

Example 2:

Input: area = 37
Output: [37,1]

Example 3:

Input: area = 122122
Output: [427,286]

Constraints:

1 <= area <= 10⁷

Step 01

Brute Force Baseline

Problem summary: A web developer needs to know how to design a web page's size. So, given a specific rectangular web page’s area, your job by now is to design a rectangular web page, whose length L and width W satisfy the following requirements: The area of the rectangular web page you designed must equal to the given target area. The width W should not be larger than the length L, which means L >= W. The difference between length L and width W should be as small as possible. Return an array [L, W] where L and W are the length and width of the web page you designed in sequence.

Baseline thinking

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

Pattern signal: Math

Example 1

Example 2

Example 3

Step 02

Core Insight

What unlocks the optimal approach

The W is always less than or equal to the square root of the area, so we start searching at sqrt(area) till we find the result.

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 #492: Construct the Rectangle
class Solution {
    public int[] constructRectangle(int area) {
        int w = (int) Math.sqrt(area);
        while (area % w != 0) {
            --w;
        }
        return new int[] {area / w, w};
    }
}

// Accepted solution for LeetCode #492: Construct the Rectangle
func constructRectangle(area int) []int {
	w := int(math.Sqrt(float64(area)))
	for area%w != 0 {
		w--
	}
	return []int{area / w, w}
}

# Accepted solution for LeetCode #492: Construct the Rectangle
class Solution:
    def constructRectangle(self, area: int) -> List[int]:
        w = int(sqrt(area))
        while area % w != 0:
            w -= 1
        return [area // w, w]

// Accepted solution for LeetCode #492: Construct the Rectangle
struct Solution;

impl Solution {
    fn construct_rectangle(area: i32) -> Vec<i32> {
        let mut max = (area as f64).sqrt().floor() as i32;
        while area % max != 0 {
            max -= 1;
        }
        vec![area / max, max]
    }
}

#[test]
fn test() {
    assert_eq!(Solution::construct_rectangle(4), vec![2, 2]);
}

// Accepted solution for LeetCode #492: Construct the Rectangle
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #492: Construct the Rectangle
// class Solution {
//     public int[] constructRectangle(int area) {
//         int w = (int) Math.sqrt(area);
//         while (area % w != 0) {
//             --w;
//         }
//         return new int[] {area / w, w};
//     }
// }

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

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.