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++
// C++ program for the above approachÂ
#include <bits/stdc++.h>using namespace std;Â
// Function to count the number of// largest circles in a rectangleint totalCircles(int L, int B){    // If length exceeds breadth    if (L > B) {Â
        // Swap to reduce length        // to smaller than breadth        int temp = L;        L = B;        B = temp;    }Â
    // Return total count    // of circles inscribed    return B / L;}Â
// Driver Codeint main(){Â Â Â Â int L = 3;Â Â Â Â int B = 8;Â Â Â Â cout << totalCircles(L, B);Â
    return 0;} |
Java
// Java program for the above approachimport java.io.*;import java.util.*;class GFG {Â
  // 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) {Â
      // Swap to reduce length      // to smaller than breadth      int temp = L;      L = B;      B = temp;    }Â
    // Return total count    // of circles inscribed    return B / L;  }Â
  // Driver Code  public static void main(String[] args)  {    int L = 3;    int B = 8;    System.out.print(totalCircles(L, B));  }}Â
// This code is contributed by susmitakundugoaldanga. |
Python3
# Python3 program for the above approachÂ
# Function to count the number of# largest circles in a rectangledef totalCircles(L, B) :         # If length exceeds breadth    if (L > B) :Â
        # Swap to reduce length        # to smaller than breadth        temp = L        L = B        B = temp         # Return total count    # of circles inscribed    return B // LÂ
# Driver CodeL = 3B = 8print(totalCircles(L, B))Â
# This code is contributed by splevel62. |
C#
// C# program to implement// the above approachusing System;public class GFG{Â
  // 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) {Â
      // Swap to reduce length      // to smaller than breadth      int temp = L;      L = B;      B = temp;    }Â
    // Return total count    // of circles inscribed    return B / L;  }Â
Â
  // Driver Code  public static void Main(String[] args)  {    int L = 3;    int B = 8;    Console.Write(totalCircles(L, B));  }}Â
// This code is contributed by souravghosh0416. |
Javascript
<script>Â
// javascript program to implement// the above approachÂ
    // 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 CodeÂ
    var L = 3;    var B = 8;    document.write(totalCircles(L, B).toString().split('.')[0]);   Â
</script> |
Output:Â
2
Â
Time Complexity: O(1)
Auxiliary Space: O(1)
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