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 Ã— 1 that can be drawn inside a bigger rectangle of dimension 5 Ã— 3.

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

Javascript

 ``

Output:

`6`

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

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Previous
Next