# Minimum area of square holding two identical rectangles

• Difficulty Level : Medium
• Last Updated : 01 Sep, 2021

Given the length L and breadth B of two identical rectangles, the task is to find the minimum area of a square in which the two identical rectangles with dimensions L × B can be embedded.
Examples:

Input: L = 7, B = 4
Output: 64
Explanation: Two rectangles with sides 7 x 4 can fit into square with side 8. By placing two rectangles with side 4 upon each other and the length of contact is 7.
Input: L = 1, B = 3
Output:
Explanation: Two rectangles with sides 1 x 3 can fit into square with side 3. By placing two rectangles with side 1 upon each other and a gap of 1 between the 2 rectangles.

Approach:

• If one side of the rectangle is lesser than or equal to half the length of the other side then the side of the square is the longer side of the rectangle.
• If twice the length of the smaller side is greater than the larger side, then the side of the square is twice the length of the smaller side of the rectangle.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above problem``#include ``using` `namespace` `std;` `// Function to find the``// area of the square``int` `areaSquare(``int` `L, ``int` `B)``{``    ``// Larger side of rectangle``    ``int` `large = max(L, B);` `    ``// Smaller side of the rectangle``    ``int` `small = min(L, B);` `    ``if` `(large >= 2 * small)``        ``return` `large * large;``    ``else``        ``return` `(2 * small) * (2 * small);``}` `// Driver code``int` `main()``{``    ``int` `L = 7;``    ``int` `B = 4;``    ``cout << areaSquare(L, B);``    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``import` `java.io.*;` `class` `GFG{` `// Function to find the``// area of the square``static` `int` `areaSquare(``int` `L, ``int` `B)``{``    ` `    ``// Larger side of rectangle``    ``int` `large = Math.max(L, B);` `    ``// Smaller side of the rectangle``    ``int` `small = Math.min(L, B);` `    ``if` `(large >= ``2` `* small)``    ``{``        ``return` `large * large;``    ``}``    ``else``    ``{``        ``return` `(``2` `* small) * (``2` `* small);``    ``}``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `L = ``7``;``    ``int` `B = ``4``;``    ` `    ``System.out.println(areaSquare(L, B));``}``}` `// This code is contributed by offbeat`

## Python3

 `# Python3 program for the above problem` `# Function to find the``# area of the square``def` `areaSquare(L, B):` `    ``# Larger side of rectangle``    ``large ``=` `max``(L, B)` `    ``# Smaller side of the rectangle``    ``small ``=` `min``(L, B)` `    ``if``(large >``=` `2` `*` `small):``        ``return` `large ``*` `large``    ``else``:``        ``return` `(``2` `*` `small) ``*` `(``2` `*` `small)``        ` `# Driver code``if` `__name__ ``=``=` `'__main__'``:` `    ``L ``=` `7``    ``B ``=` `4``    ` `    ``print``(areaSquare(L, B))` `# This code is contributed by Shivam Singh`

## C#

 `// C# program for the above problem``using` `System;` `class` `GFG{` `// Function to find the``// area of the square``public` `static` `int` `areaSquare(``int` `L, ``int` `B)``{``    ` `    ``// Larger side of rectangle``    ``int` `large = Math.Max(L, B);` `    ``// Smaller side of the rectangle``    ``int` `small = Math.Min(L, B);` `    ``if` `(large >= 2 * small)``    ``{``        ``return` `large * large;``    ``}``    ``else``    ``{``        ``return` `(2 * small) * (2 * small);``    ``}``}` `// Driver code``public` `static` `void` `Main()``{``    ``int` `L = 7;``    ``int` `B = 4;``    ` `    ``Console.Write(areaSquare(L, B));``}``}` `// This code is contributed by Code_Mech`

## Javascript

 ``

Output:

`64`

Time Complexity: O(1)
Auxiliary Space Complexity: O(1)

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