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# Maximum given sized rectangles that can be cut out of a sheet of paper

Given the length L and breadth B of a sheet of paper, the task is to find the maximum number of rectangles with given length l and breadth b that can be cut from this sheet of paper.
Examples:

Input: L = 5, B = 2, l = 14, b = 3
Output:
The sheet is smaller than the required rectangle. So, no rectangle of the given dimension can be cut from the sheet.
Input: L = 10, B = 7, l = 4, b = 3
Output:

Approach:

• Try to cut the rectangles horizontally i.e. length of the rectangle is aligned with the length of the sheet and breadth of the rectangle is aligned with the breadth of the sheet and store the count of rectangles possible in horizontal.
• Repeat the same with vertical alignment i.e. when length of the rectangle is aligned with the breadth of the sheet and breadth of the rectangle is aligned with the length of the sheet and store the result in vertical. Print max(horizontal, vertical) as the result.

Below is the implementation of the above approach:

## C++

 `// CPP implementation of the approach` `#include` `using` `namespace` `std;`   `// Function to return the maximum rectangles possible` `int` `maxRectangles(``int` `L, ``int` `B, ``int` `l, ``int` `b)` `{` `    ``int` `horizontal = 0, vertical = 0;`   `    ``// Cut rectangles horizontally if possible` `    ``if` `(l <= L && b <= B) ` `    ``{`   `        ``// One rectangle is a single cell` `        ``int` `columns = B / b;` `        ``int` `rows = L / l;`   `        ``// Total rectangles = total cells` `        ``horizontal = rows * columns;` `    ``}`   `    ``// Cut rectangles vertically if possible` `    ``if` `(l <= B && b <= L)` `    ``{` `        ``int` `columns = L / b;` `        ``int` `rows = B / l;`   `        ``vertical = rows * columns;` `    ``}`   `    ``// Return the maximum possible rectangles` `    ``return` `max(horizontal, vertical);` `}`   `// Driver code` `int` `main()` `{` `    ``int` `L = 10, B = 7, l = 4, b = 3;` `    ``cout << (maxRectangles(L, B, l, b)) << endl;` `}`   `// This code is contributed by` `// Sanjit_Prasad`

## Java

 `// Java implementation of the approach` `class` `GFG {`   `    ``// Function to return the maximum rectangles possible` `    ``static` `int` `maxRectangles(``int` `L, ``int` `B, ``int` `l, ``int` `b)` `    ``{` `        ``int` `horizontal = ``0``, vertical = ``0``;`   `        ``// Cut rectangles horizontally if possible` `        ``if` `(l <= L && b <= B) {`   `            ``// One rectangle is a single cell` `            ``int` `columns = B / b;` `            ``int` `rows = L / l;`   `            ``// Total rectangles = total cells` `            ``horizontal = rows * columns;` `        ``}`   `        ``// Cut rectangles vertically if possible` `        ``if` `(l <= B && b <= L) {` `            ``int` `columns = L / b;` `            ``int` `rows = B / l;`   `            ``vertical = rows * columns;` `        ``}`   `        ``// Return the maximum possible rectangles` `        ``return` `Math.max(horizontal, vertical);` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `L = ``10``, B = ``7``, l = ``4``, b = ``3``;` `        ``System.out.print(maxRectangles(L, B, l, b));` `    ``}` `}`

## Python3

 `# Python3 implementation of the approach `   `# Function to return the maximum ` `# rectangles possible ` `def` `maxRectangles(L, B, l, b): `   `    ``horizontal, vertical ``=` `0``, ``0`   `    ``# Cut rectangles horizontally if possible ` `    ``if` `l <``=` `L ``and` `b <``=` `B:`   `        ``# One rectangle is a single cell ` `        ``columns ``=` `B ``/``/` `b ` `        ``rows ``=` `L ``/``/` `l `   `        ``# Total rectangles = total cells ` `        ``horizontal ``=` `rows ``*` `columns `   `    ``# Cut rectangles vertically if possible ` `    ``if` `l <``=` `B ``and` `b <``=` `L:` `    `  `        ``columns ``=` `L ``/``/` `b ` `        ``rows ``=` `B ``/``/` `l `   `        ``vertical ``=` `rows ``*` `columns `   `    ``# Return the maximum possible rectangles ` `    ``return` `max``(horizontal, vertical) `   `# Driver code ` `if` `__name__ ``=``=` `"__main__"``:`   `    ``L, B, l, b ``=` `10``, ``7``, ``4``, ``3` `    ``print``(maxRectangles(L, B, l, b)) `   `# This code is contributed by Rituraj Jain`

## C#

 `// C# implementation of the above approach ` `using` `System;`   `class` `GFG ` `{ `   `    ``// Function to return the ` `    ``// maximum rectangles possible ` `    ``static` `int` `maxRectangles(``int` `L, ``int` `B, ` `                                ``int` `l, ``int` `b) ` `    ``{ ` `        ``int` `horizontal = 0, vertical = 0; `   `        ``// Cut rectangles horizontally if possible ` `        ``if` `(l <= L && b <= B)` `        ``{ `   `            ``// One rectangle is a single cell ` `            ``int` `columns = B / b; ` `            ``int` `rows = L / l; `   `            ``// Total rectangles = total cells ` `            ``horizontal = rows * columns; ` `        ``} `   `        ``// Cut rectangles vertically if possible ` `        ``if` `(l <= B && b <= L) ` `        ``{ ` `            ``int` `columns = L / b; ` `            ``int` `rows = B / l; ` `            ``vertical = rows * columns; ` `        ``} `   `        ``// Return the maximum possible rectangles ` `        ``return` `Math.Max(horizontal, vertical); ` `    ``} `   `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `L = 10, B = 7, l = 4, b = 3; ` `        ``Console.WriteLine(maxRectangles(L, B, l, b)); ` `    ``} ` `} `   `// This code is contributed by Ryuga`

## PHP

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## Javascript

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Output:

`4`

Time Complexity: O(1), since there is only a basic arithmetic operation that takes constant time.
Auxiliary Space: O(1), since no extra space has been taken.

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