Given two integers L and B representing the length and breadth of a rectangle, the task is to find the maximum number of largest possible circles that can be inscribed in the given rectangle without overlapping.
Examples:
Input: L = 3, B = 8
Output: 2
Explanation:From the above figure it can be clearly seen that the largest circle with a diameter of 3 cm can be inscribed in the given rectangle.
Therefore, the count of such circles is 2.Input: L = 2, B = 9
Output: 4
Approach: The given problem can be solved based on the following observations:
- The largest circle that can be inscribed in a rectangle will have diameter equal to the smaller side of the rectangle.
- Therefore, the maximum number of such largest circles possible is equal to ( Length of the largest side ) / ( Length of the smallest side ).
Therefore, from the above observation, simply print the value of ( Length of the largest side ) / ( Length of the smallest side ) as the required result.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to count the number of // largest circles in a rectangle int totalCircles( int L, int B)
{ // If length exceeds breadth
if (L > B) {
// Swap to reduce length
// to smaller than breadth
int temp = L;
L = B;
B = temp;
}
// Return total count
// of circles inscribed
return B / L;
} // Driver Code int main()
{ int L = 3;
int B = 8;
cout << totalCircles(L, B);
return 0;
} |
// Java program for the above approach import java.io.*;
import java.util.*;
class GFG
{ // Function to count the number of
// largest circles in a rectangle
static int totalCircles( int L, int B)
{
// If length exceeds breadth
if (L > B) {
// Swap to reduce length
// to smaller than breadth
int temp = L;
L = B;
B = temp;
}
// Return total count
// of circles inscribed
return B / L;
}
// Driver Code
public static void main(String[] args)
{
int L = 3 ;
int B = 8 ;
System.out.print(totalCircles(L, B));
}
} // This code is contributed by susmitakundugoaldanga. |
# Python3 program for the above approach # Function to count the number of # largest circles in a rectangle def totalCircles(L, B) :
# If length exceeds breadth
if (L > B) :
# Swap to reduce length
# to smaller than breadth
temp = L
L = B
B = temp
# Return total count
# of circles inscribed
return B / / L
# Driver Code L = 3
B = 8
print (totalCircles(L, B))
# This code is contributed by splevel62. |
// C# program to implement // the above approach using System;
public class GFG
{ // Function to count the number of
// largest circles in a rectangle
static int totalCircles( int L, int B)
{
// If length exceeds breadth
if (L > B) {
// Swap to reduce length
// to smaller than breadth
int temp = L;
L = B;
B = temp;
}
// Return total count
// of circles inscribed
return B / L;
}
// Driver Code
public static void Main(String[] args)
{
int L = 3;
int B = 8;
Console.Write(totalCircles(L, B));
}
} // This code is contributed by souravghosh0416. |
<script> // javascript program to implement // the above approach // Function to count the number of
// largest circles in a rectangle
function totalCircles( L, B)
{
// If length exceeds breadth
if (L > B) {
// Swap to reduce length
// to smaller than breadth
var temp = L;
L = B;
B = temp;
}
// Return total count
// of circles inscribed
return B / L;
}
// Driver Code
var L = 3;
var B = 8;
document.write(totalCircles(L, B).toString().split( '.' )[0]);
</script> |
2
Time Complexity: O(1)
Auxiliary Space: O(1)