Minimum squares to cover a rectangle

Given a rectangle with length l and breadth b, we need to find the minimum number of squares that can cover the surface of the rectangle, given that each square has a side of length a. It is allowed to cover the surface larger than the rectangle, but the rectangle has to be covered. It is not allowed to break the square.

Examples:

Input : 1 2 3
Output :1
We have a 3x3 square and we need
to make a rectangle of size 1x2.
So we need only 1 square to cover the
rectangle.

Input : 11 23 14
Output :2 

The only way to actually fill the rectangle optimally is to arrange each square such that it is parallel to the sides of the rectangle. So we just need to find the number of squares to fully cover the length and breadth of the rectangle. 
The length of the rectangle is l, and if the side length of the square is a divided l, then there must be l/a squares to cover the full length of l. If l isn’t divisible by a, we need to add 1 to l/a, to round it down. For this, we can use the ceil function, as ceil(x) returns the least integer which is above or equal to x. 

We can do the same with the rectangle width and take the number of squares across the width to be ceil(b/a). 

So, total number of squares=ceil(m/a) * ceil(n/a). 

C++

// C++ program to find the minimum number
// of squares to cover the surface of the
// rectangle with given dimensions
#include <bits/stdc++.h>
using namespace std;
int squares(int l, int b, int a)
{
    // function to count
    // the number of squares that can
    // cover the surface of the rectangle
    return ceil(l / (double)a) * ceil(b / (double)a);
}

// Driver code
int main()
{
    int l = 11, b = 23, a = 14;
    cout << squares(l, b, a) << endl;
    return 0;
}

Java

// Java program to find the minimum number
// of squares to cover the surface of the
// rectangle with given dimensions
class GFG
{
static int squares(int l, int b, int a)
{

// function to count
// the number of squares that can
// cover the surface of the rectangle
return (int)(Math.ceil(l / (double)a) *
             Math.ceil(b / (double)a));
}

// Driver code
public static void main(String[] args)
{
    int l = 11, b = 23, a = 14;
    System.out.println(squares(l, b, a));
}
}

// This code is contributed by ChitraNayal

Python 3

# Python3 program to find the minimum number
# of squares to cover the surface of the
# rectangle with given dimensions
import math

def squares(l, b, a):

    # function to count
    # the number of squares that can
    # cover the surface of the rectangle
    return math.ceil(l / a) * math.ceil(b / a)

# Driver code
if __name__ == "__main__":
    l = 11
    b = 23
    a = 14
    print(squares(l, b, a))

# This code is contributed
# by ChitraNayal

C#

// C# program to find the minimum number
// of squares to cover the surface of the
// rectangle with given dimensions
using System;

class GFG
{
static int squares(int l, int b, int a)
{

// function to count
// the number of squares that can
// cover the surface of the rectangle
return (int)(Math.Ceiling(l / (double)a) *
             Math.Ceiling(b / (double)a));
}

// Driver code
public static void Main()
{
    int l = 11, b = 23, a = 14;
    Console.Write(squares(l, b, a));
}
}

// This code is contributed by ChitraNayal

PHP

<?php
// PHP program to find the minimum number
// of squares to cover the surface of the
// rectangle with given dimensions

function squares($l, $b, $a)
{
    // function to count
    // the number of squares that can
    // cover the surface of the rectangle
    return ceil($l / (double)$a) *
           ceil($b / (double)$a);
}

// Driver code
$l = 11;
$b = 23;
$a = 14;
echo squares($l, $b, $a);

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

Javascript

<script>

// javascript program to find the minimum number
// of squares to cover the surface of the
// rectangle with given dimensions

function squares(l , b , a)
{

    // function to count
    // the number of squares that can
    // cover the surface of the rectangle
    return parseInt(Math.ceil(l / a) *
                 Math.ceil(b / a));
}

// Driver code

var l = 11, b = 23, a = 14;
document.write(squares(l, b, a));

// This code is contributed by Amit Katiyar

</script>

Output:

2

Time complexity: O(1), since there is no loop or recursion.

Auxiliary Space: O(1), since no extra space has been taken.