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

Last Updated : 23 Jun, 2022

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: 0
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: 4

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<bits/stdc++.h>
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

<?php
// PHP implementation of the approach
 
// Function to return the maximum 
// rectangles possible
function maxRectangles($L, $B, $l, $b)
{
    $horizontal = 0;
    $vertical = 0;
 
    // Cut rectangles horizontally if possible
    if ($l <= $L && $b <= $B) 
    {
 
        // One rectangle is a single cell
        $columns = (int)($B / $b);
        $rows = (int)($L / $l);
 
        // Total rectangles = total cells
        $horizontal = $rows * $columns;
    }
 
    // Cut rectangles vertically if possible
    if ($l <= $B && $b <= $L)
    {
        $columns = (int)($L / $b);
        $rows = (int)($B / $l);
 
        $vertical = $rows * $columns;
    }
 
    // Return the maximum possible rectangles
    return max($horizontal, $vertical);
}
 
// Driver code
$L = 10;
$B = 7;
$l = 4;
$b = 3;
print(maxRectangles($L, $B, $l, $b));
 
// This code is contributed by mits
?>

Javascript

<script>
 
// Javascript implementation of the approach
 
// Function to return the maximum 
// rectangles possible
function maxRectangles(L, B, l, b)
{
    var horizontal = 0, vertical = 0;
 
    // Cut rectangles horizontally 
    // if possible
    if (l <= L && b <= B) 
    {
         
        // One rectangle is a single cell
        var columns = parseInt(B / b);
        var rows = parseInt(L / l);
 
        // Total rectangles = total cells
        horizontal = rows * columns;
    }
 
    // Cut rectangles vertically if possible
    if (l <= B && b <= L) 
    {
        var columns = parseInt(L / b);
        var rows = parseInt(B / l);
 
        vertical = rows * columns;
    }
 
    // Return the maximum possible rectangles
    return Math.max(horizontal, vertical);
}
 
// Driver Code
var L = 10, B = 7, l = 4, b = 3;
 
document.write(maxRectangles(L, B, l, b));
 
// This code is contributed by kirti
 
</script>

Output:

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.