# 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++

 `// C++ program for the above approach  ` ` `  `#include   ` `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;  ` `} `

## Java

 `// 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 `

## Python3

 `# 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 `

## C#

 `// 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 `

Output:

```6
```

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

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Improved By : jana_sayantan