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: 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.
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 squareint 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 codeint main(){ int L = 7; int B = 4; cout << areaSquare(L, B); return 0;} |
Java
// Java implementation of the above approachimport java.io.*;class GFG{// Function to find the// area of the squarestatic 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 codepublic 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 squaredef 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 codeif __name__ == '__main__': L = 7 B = 4 print(areaSquare(L, B))# This code is contributed by Shivam Singh |
C#
// C# program for the above problemusing System;class GFG{// Function to find the// area of the squarepublic 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 codepublic 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 squarefunction 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)
Auxiliary Space Complexity: O(1)
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