Nos dan la suma de la longitud, la anchura y la altura , digamos S, de un paralelepípedo. La tarea es encontrar el volumen máximo que se puede lograr para que la suma de los lados sea S.
Volumen de un paralelepípedo = largo * ancho * alto
Ejemplos:
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
Método 1: (Fuerza bruta)
La idea de ejecutar tres anidados, uno para largo, uno para ancho y otro para alto. Para cada iteración, calcule el volumen y compárelo con el volumen máximo.
A continuación se muestra la implementación de este enfoque:
C++
#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
// 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
# 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#
// 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.
PHP
<?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.
?>
Javascript
<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>
Producción :
18
Complejidad de tiempo: O(n 2 )
Espacio auxiliar: O(1)
Método 2: (Enfoque eficiente)
La idea es dividir los bordes lo más equitativamente posible.
Entonces,
largo = piso(s/3)
ancho = piso((s – largo)/2) = piso((s – piso(s/3)/2)
alto = s – largo – ancho = s – piso(s /3) – piso((s – piso(s/3))/2)
A continuación se muestra la implementación de este enfoque:
C++
#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
// 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
# 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#
// 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.
PHP
<?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.
?>
Javascript
<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>
Producción :
18
Complejidad de Tiempo: O(1)
Espacio Auxiliar: O(1)
¿Cómo funciona esto?
Básicamente necesitamos maximizar el producto de
tres números, x, y y z cuya suma está dada.
Dado s = x + y + z
Maximizar P = x * y * z
= x * y * (s – x – y)
= x*y*s – x*x*s – x*y*y
Obtenemos dp/ dx = sy – 2xy – y*y
y dp/dy = sx – 2xy – x*x
Obtenemos dp/dx = 0 y dp/dy = 0 cuando
x = s/3, y = s/3
Entonces z = s – x – y = s/3
Sugiera si alguien tiene una mejor solución que sea más eficiente en términos de espacio y tiempo.
Este artículo es una contribución de Aarti_Rathi . Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.