容纳两个相同矩形的正方形的最小面积

给定两个相同矩形的长度L和宽度B ，任务是找到正方形的最小面积，在其中可以嵌入尺寸为L×B的两个相同矩形。

例子：

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: 9
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.

方法：

• 如果矩形的一侧小于或等于另一侧的一半，则正方形的一侧为矩形的较长侧。
• 如果较小边的长度的两倍大于较大边的长度，则正方形的边是矩形较小边的长度的两倍。

下面是上述方法的实现：

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

输出：

64

时间复杂度： O(1)
辅助空间复杂度： O(1)