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Minimum area of square holding two identical rectangles
  • Difficulty Level : Hard
  • Last Updated : 14 May, 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 <bits/stdc++.h>
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




<script>
 
// Javascript implementation of the above approach
 
// Function to find the
// area of the square
function areaSquare(L, B)
{
       
    // Larger side of rectangle
    let large = Math.max(L, B);
   
    // Smaller side of the rectangle
    let small = Math.min(L, B);
   
    if (large >= 2 * small)
    {
        return large * large;
    }
    else
    {
        return (2 * small) * (2 * small);
    }
}
  
  // Driver Code
     
    let L = 7;
    let B = 4;
       
    document.write(areaSquare(L, B));
      
</script>
Output: 
64

 

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

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