Construct the Rectangle

Construct the Rectangle - Problem

Math Easy

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.

Input & Output

Example 1 — Perfect Square Area

$ Input: area = 4

› Output: [2,2]

💡 Note: Area 4 can form a 2×2 square, which has the minimum possible difference of 0

Example 2 — Non-Square Area

$ Input: area = 6

› Output: [3,2]

💡 Note: Area 6 has factors: 1×6 (diff=5) and 2×3 (diff=1). The 3×2 rectangle has minimum difference

Example 3 — Prime Number Area

$ Input: area = 7

› Output: [7,1]

💡 Note: Area 7 is prime, so only possible rectangle is 7×1

Constraints

1 ≤ area ≤ 10⁷

Visualization

Tap to expand

Asked in

G Google 15 a Amazon 12

Brief HTML summary: The key insight is that the optimal rectangle is closest to a square. Best approach starts from √area and works backwards: Time: O(√n), Space: O(1)

Common Approaches

✓ Dfs

⏱️ Time: N/A Space: N/A

Brute Force - Linear Search

⏱️ Time: O(n) Space: O(1)

Iterate through all possible widths from 1 to area, and for each width that divides the area evenly, calculate the corresponding length. Keep track of the pair with minimum difference.

Optimized - Square Root Approach

⏱️ Time: O(√n) Space: O(1)

Since we want the rectangle closest to a square, we start from the square root of the area and work backwards. The first width we find that divides the area evenly will give us the optimal rectangle with minimum difference.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>

int main() {
    int area;
    scanf("%d", &area);
    
    int W = 1;
    
    // Find the largest factor W such that W * W <= area
    // This ensures W <= sqrt(area) and minimizes L - W
    for (int i = 1; i * i <= area; i++) {
        if (area % i == 0) {
            W = i;
        }
    }
    
    int L = area / W;
    
    printf("[%d,%d]\n", L, W);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Ln 1, Col 1

Smart Actions

💡 Explanation

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