Minimum squares to evenly cut a rectangle

Given a rectangular sheet of length l and width w. we need to divide this sheet into square sheets such that the number of square sheets should be as minimum as possible.
Examples:

Input :l= 4 w=6 
Output :6 
We can form squares with side of 1 unit, But the number of squares will be 24, this is not minimum. If we make square with side of 2, then we have 6 squares. and this is our required answer. 
And also we can’t make square with side 3, if we select 3 as square side, then whole sheet can’t be converted into squares of equal length. 

Input :l=3 w=5 
Output :15

Optimal length of the side of a square is equal to GCD of two numbers

C++

// CPP program to find minimum number of
// squares to make a given rectangle.
#include <bits/stdc++.h>
using namespace std;

int countRectangles(int l, int w)
{
    // if we take gcd(l, w), this
    // will be largest possible
    // side for square, hence minimum
    // number of square.
    int squareSide = __gcd(l, w);

    // Number of squares.
    return (l * w) / (squareSide * squareSide);
}

// Driver code
int main()
{
    int l = 4, w = 6;
    cout << countRectangles(l, w) << endl;
    return 0;
}

Java

// Java program to find minimum number of
// squares to make a given rectangle.

class GFG{
static int __gcd(int a, int b) {
   if (b==0) return a;
   return __gcd(b,a%b);
}
static int countRectangles(int l, int w)
{
    // if we take gcd(l, w), this
    // will be largest possible
    // side for square, hence minimum
    // number of square.
    int squareSide = __gcd(l, w);

    // Number of squares.
    return (l * w) / (squareSide * squareSide);
}

// Driver code
public static void main(String[] args)
{
    int l = 4, w = 6;
    System.out.println(countRectangles(l, w));
}
}
// This code is contributed by mits

Python3

# Python3 code to find minimum number of
# squares to make a given rectangle.

import math

def countRectangles(l, w):

    # if we take gcd(l, w), this
    # will be largest possible
    # side for square, hence minimum
    # number of square.
    squareSide = math.gcd(l,w)

    # Number of squares.
    return (l*w)/(squareSide*squareSide)

# Driver Code

if __name__ == '__main__':
    l = 4
    w = 6
    ans = countRectangles(l, w)
    print (int(ans))

# this code is contributed by
# SURENDRA_GANGWAR

C#

// C# program to find minimum number of
// squares to make a given rectangle.

class GFG{
static int __gcd(int a, int b) {
if (b==0) return a;
return __gcd(b,a%b);
}
static int countRectangles(int l, int w)
{
    // if we take gcd(l, w), this
    // will be largest possible
    // side for square, hence minimum
    // number of square.
    int squareSide = __gcd(l, w);

    // Number of squares.
    return (l * w) / (squareSide * squareSide);
}

// Driver code
public static void Main()
{
    int l = 4, w = 6;
    System.Console.WriteLine(countRectangles(l, w));
}
}
// This code is contributed by mits

PHP

<?php
// PHP program to find minimum number
// of squares to make a given rectangle.

function gcd($a, $b)
{
    return $b ? gcd($b, $a % $b) : $a;
}

function countRectangles($l, $w)
{
    // if we take gcd(l, w), this
    // will be largest possible
    // side for square, hence minimum
    // number of square.
    $squareSide = gcd($l, $w);

    // Number of squares.
    return ($l * $w) / ($squareSide *
                        $squareSide);
}

// Driver code
$l = 4;
$w = 6;
echo countRectangles($l, $w) . "\n";

// This code is contributed
// by ChitraNayal
?>

Javascript

<script>

// Javascript program to find minimum number of
// squares to make a given rectangle.

function __gcd(a, b) {
    if (b==0) return a;
    return __gcd(b,a%b);
}

function countRectangles(l, w)
{
    // if we take gcd(l, w), this
    // will be largest possible
    // side for square, hence minimum
    // number of square.
    let squareSide = __gcd(l, w);

    // Number of squares.
    return parseInt((l * w) / (squareSide * squareSide));
}

// Driver code
    let l = 4, w = 6;
    document.write(countRectangles(l, w));

</script>

Output:

6

Time Complexity: O(log(min(a, b))), where a and b are two parameters of gcd.

Auxiliary Space: O(log(min(a, b)))