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 ，任务是找到可以嵌入尺寸为L × B的两个相同矩形的正方形的最小面积。
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：

• 如果矩形的一边小于或等于另一边长度的一半，则正方形的边是矩形的长边。
• 如果小边长度的两倍大于大边的长度，则正方形的边是矩形小边长度的两倍。

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

Javascript

 
// 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));
      

输出：

64

时间复杂度： O(1)
辅助空间复杂度： O(1)

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。