Count of smaller rectangles that can be placed inside a bigger rectangle

Given four integers L, B, l, and b, where L and B denote the dimensions of a bigger rectangle and l and b denotes the dimension of a smaller rectangle, the task is to count the number of smaller rectangles that can be drawn inside a bigger rectangle. 
Note: Smaller rectangles can overlap partially.

Examples:

Input: L = 5, B = 3, l = 4, b = 1
Output: 6
Explanation:
There are 6 rectangles of dimension 4 × 1 that can be drawn inside a bigger rectangle of dimension 5 × 3.

Input: L = 3, B = 2, l = 2, b = 1
Output: 3
Explanation:
There are 3 rectangles of dimension 3 × 2 can be drawn inside a bigger rectangle of dimension 2 × 1.

Naive Approach: The idea is to iterate over the length L and breadth B of the bigger rectangle to count the number of smaller rectangles of dimension l x b that can be drawn within the range of bigger rectangle. Print the total count after the traversal. 

Time Complexity: O(L * B)
Auxiliary Space: O(1)

Efficient Approach: The above problem can be solved using Permutation and Combinations. Below are the steps:



  1. The total possible values of the length of smaller rectangle l using the length L is given by (L – l + 1).
  2. The total possible values of the breadth of smaller rectangle b using the length B is given by (B – b + 1).
  3. Hence, the total number of possible rectangles can be formed is given by:

    (L – l + 1) * (B – b + 1)

Below is the implementation of the above approach:

C++

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// C++ program for the above approach 
  
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to count smaller rectangles 
// within the larger rectangle 
int No_of_rectangles(int L, int B, 
                    int l, int b) 
    // If the dimension of the smaller 
    // rectangle is greater than the 
    // bigger one 
    if ((l > L) || (b > B)) { 
        return -1; 
    
  
    else
  
        // Return the number of smaller 
        // rectangles possible 
        return (L - l + 1) * (B - b + 1); 
    
  
// Driver Code 
int main() 
    // Dimension of bigger rectangle 
    int L = 5, B = 3; 
  
    // Dimension of smaller rectangle 
    int l = 4, b = 1; 
  
    // Function call 
    cout << No_of_rectangles(L, B, l, b); 
  
    return 0; 
}

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Java

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// Java program for the above approach
import java.util.*;
  
class GFG{
  
// Function to count smaller rectangles
// within the larger rectangle
static int No_of_rectangles(int L, int B,
                            int l, int b)
{
      
    // If the dimension of the smaller
    // rectangle is greater than the
    // bigger one
    if ((l > L) || (b > B)) 
    {
        return -1;
    }
  
    else
    {
          
        // Return the number of smaller
        // rectangles possible
        return (L - l + 1) * (B - b + 1);
    }
}
  
// Driver Code
public static void main(String[] args)
{
      
    // Dimension of bigger rectangle
    int L = 5, B = 3;
  
    // Dimension of smaller rectangle
    int l = 4, b = 1;
  
    // Function call
    System.out.println(No_of_rectangles(L, B, l, b));
}
}
  
// This code is contributed by jana_sayantan

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Python3

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# Python3 program for the above approach
  
# Function to count smaller rectangles
# within the larger rectangle
def No_of_rectangles( L, B, l, b):
  
    # If the dimension of the smaller
    # rectangle is greater than the
    # bigger one
    if (l > L) or (b > B): 
        return -1;
          
    else:
  
        # Return the number of smaller
        # rectangles possible
        return (L - l + 1) * (B - b + 1);
      
# Driver code
if __name__ == '__main__'
      
    # Dimension of bigger rectangle
    L = 5
    B = 3
  
    # Dimension of smaller rectangle
    l = 4
    b = 1
  
    # Function call
    print(No_of_rectangles(L, B, l, b))
  
# This code is contributed by jana_sayantan

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C#

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// C# program for the above approach
using System;
  
class GFG{
  
// Function to count smaller rectangles
// within the larger rectangle
static int No_of_rectangles(int L, int B,
                            int l, int b)
{
      
    // If the dimension of the smaller
    // rectangle is greater than the
    // bigger one
    if ((l > L) || (b > B))
    {
        return -1;
    }
    else
    {
          
        // Return the number of smaller
        // rectangles possible
        return (L - l + 1) * (B - b + 1);
    }
}
  
// Driver Code
public static void Main(String[] args)
{
      
    // Dimension of bigger rectangle
    int L = 5, B = 3;
      
    // Dimension of smaller rectangle
    int l = 4, b = 1;
      
    // Function call
    Console.Write(No_of_rectangles(L, B, l, b));
}
}
  
// This code is contributed by jana_sayantan

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Output: 

6

Time Complexity: O(1)
Auxiliary Space: O(1)

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Improved By : jana_sayantan