We are given the sum of length, breadth and height, say S, of a cuboid. The task is to find the maximum volume that can be achieved so that sum of side is S.
Volume of a cuboid = length * breadth * height
Examples :
Input : s = 4
Output : 2
Only possible dimensions are some combination of 1, 1, 2.
Input : s = 8
Output : 18
All possible edge dimensions:
[1, 1, 6], volume = 6
[1, 2, 5], volume = 10
[1, 3, 4], volume = 12
[2, 2, 4], volume = 16
[2, 3, 3], volume = 18
Method 1: (Brute Force)
The idea to run three nested, one for length, one for breadth and one for height. For each iteration, calculate the volume and compare with maximum volume.
Below is the implementation of this approach:
#include <bits/stdc++.h> using namespace std;
// Return the maximum volume. int maxvolume( int s)
{ int maxvalue = 0;
// for length
for ( int i = 1; i <= s - 2; i++) {
// for breadth
for ( int j = 1; j <= s - 1; j++) {
// for height
int k = s - i - j;
// calculating maximum volume.
maxvalue = max(maxvalue, i * j * k);
}
}
return maxvalue;
} // Driven Program int main()
{ int s = 8;
cout << maxvolume(s) << endl;
return 0;
} |
// Java code to Maximize volume of // cuboid with given sum of sides class GFG
{ // Return the maximum volume.
static int maxvolume( int s)
{
int maxvalue = 0 ;
// for length
for ( int i = 1 ; i <= s - 2 ; i++)
{
// for breadth
for ( int j = 1 ; j <= s - 1 ; j++)
{
// for height
int k = s - i - j;
// calculating maximum volume.
maxvalue = Math.max(maxvalue, i * j * k);
}
}
return maxvalue;
}
// Driver function
public static void main (String[] args)
{
int s = 8 ;
System.out.println(maxvolume(s));
}
} // This code is contributed by Anant Agarwal. |
# Python3 code to Maximize volume of # cuboid with given sum of sides # Return the maximum volume. def maxvolume (s):
maxvalue = 0
# for length
i = 1
for i in range (s - 1 ):
j = 1
# for breadth
for j in range (s):
# for height
k = s - i - j
# calculating maximum volume.
maxvalue = max (maxvalue, i * j * k)
return maxvalue
# Driven Program s = 8
print (maxvolume(s))
# This code is contributed by "Sharad_Bhardwaj". |
// C# code to Maximize volume of // cuboid with given sum of sides using System;
class GFG
{ // Return the maximum volume.
static int maxvolume( int s)
{
int maxvalue = 0;
// for length
for ( int i = 1; i <= s - 2; i++)
{
// for breadth
for ( int j = 1; j <= s - 1; j++)
{
// for height
int k = s - i - j;
// calculating maximum volume.
maxvalue = Math.Max(maxvalue, i * j * k);
}
}
return maxvalue;
}
// Driver function
public static void Main ()
{
int s = 8;
Console.WriteLine(maxvolume(s));
}
} // This code is contributed by vt_m. |
<script> // javascript code to Maximize volume of // cuboid with given sum of sides // Return the maximum volume. function maxvolume( s)
{ let maxvalue = 0;
// for length
for (let i = 1; i <= s - 2; i++)
{
// for breadth
for (let j = 1; j <= s - 1; j++)
{
// for height
let k = s - i - j;
// calculating maximum volume.
maxvalue = Math.max(maxvalue, i * j * k);
}
}
return maxvalue;
} // Driver code let s = 8;
document.write(maxvolume(s));
// This code is contributed by gauravrajput1
</script> |
<?php // PHP code to Maximize volume of // cuboid with given sum of sides // Return the maximum volume. function maxvolume( $s )
{ $maxvalue = 0;
// for length
for ( $i = 1; $i <= $s - 2; $i ++)
{
// for breadth
for ( $j = 1; $j <= $s - 1; $j ++)
{
// for height
$k = $s - $i - $j ;
// calculating maximum volume.
$maxvalue = max( $maxvalue ,
$i * $j * $k );
}
}
return $maxvalue ;
} // Driver Code $s = 8;
echo (maxvolume( $s ));
// This code is contributed by vt_m. ?> |
Output :
18
Time Complexity: O(n2)
Auxiliary Space: O(1)
Method 2: (Efficient approach)
The idea is to divide edges as equally as possible.
So,
length = floor(s/3)
width = floor((s – length)/2) = floor((s – floor(s/3)/2)
height = s – length – width = s – floor(s/3) – floor((s – floor(s/3))/2)
Below is the implementation of this approach:
#include <bits/stdc++.h> using namespace std;
// Return the maximum volume. int maxvolume( int s)
{ // finding length
int length = s / 3;
s -= length;
// finding breadth
int breadth = s / 2;
// finding height
int height = s - breadth;
return length * breadth * height;
} // Driven Program int main()
{ int s = 8;
cout << maxvolume(s) << endl;
return 0;
} |
// Java code to Maximize volume of // cuboid with given sum of sides import java.io.*;
class GFG
{ // Return the maximum volume.
static int maxvolume( int s)
{
// finding length
int length = s / 3 ;
s -= length;
// finding breadth
int breadth = s / 2 ;
// finding height
int height = s - breadth;
return length * breadth * height;
}
// Driven Program
public static void main (String[] args)
{
int s = 8 ;
System.out.println ( maxvolume(s));
}
} // This code is contributed by vt_m. |
# Python3 code to Maximize volume of # cuboid with given sum of sides # Return the maximum volume. def maxvolume( s ):
# finding length
length = int (s / 3 )
s - = length
# finding breadth
breadth = s / 2
# finding height
height = s - breadth
return int (length * breadth * height)
# Driven Program s = 8
print ( maxvolume(s) )
# This code is contributed by "Sharad_Bhardwaj". |
// C# code to Maximize volume of // cuboid with given sum of sides using System;
class GFG
{ // Return the maximum volume.
static int maxvolume( int s)
{
// finding length
int length = s / 3;
s -= length;
// finding breadth
int breadth = s / 2;
// finding height
int height = s - breadth;
return length * breadth * height;
}
// Driven Program
public static void Main ()
{
int s = 8;
Console.WriteLine( maxvolume(s));
}
} // This code is contributed by vt_m. |
<script> // Return the maximum volume. function maxvolume( s)
{ // finding length
let length = parseInt(s / 3);
s -= length;
// finding breadth
let breadth = parseInt(s / 2);
// finding height
let height = s - breadth;
return length * breadth * height;
} // Driven Program let s = 8; document.write(maxvolume(s));
// This code is contributed by aashish1995 </script> |
<?php // Return the maximum volume. function maxvolume( $s )
{ // finding length
$length = (int)( $s / 3);
$s -= $length ;
// finding breadth
$breadth = (int)( $s / 2);
// finding height
$height = $s - $breadth ;
return $length * $breadth * $height ;
} // Driven Code $s = 8;
echo (maxvolume( $s ));
// This code is contributed by Ajit. ?> |
Output :
18
Time Complexity: O(1)
Auxiliary Space: O(1)
How does this work?
We basically need to maximize product of
three numbers, x, y and z whose sum is given.
Given s = x + y + z
Maximize P = x * y * z
= x * y * (s – x – y)
= x*y*s – x*x*s – x*y*y
We get dp/dx = sy – 2xy – y*y
and dp/dy = sx – 2xy – x*x
We get dp/dx = 0 and dp/dy = 0 when
x = s/3, y = s/3
So z = s – x – y = s/3
Please suggest if someone has a better solution which is more efficient in terms of space and time.