Dada la cuenta de dígitos 1, 2, 3, 4, encuentre la suma máxima posible

Dada la cuenta de los dígitos 1, 2, 3, 4. Usando estos dígitos, solo puede formar los números 234 y 12. La tarea es encontrar la suma máxima posible que se puede obtener después de formar los números. 

Nota : El objetivo es solo maximizar la suma, incluso si algunos de los dígitos quedan sin usar.

Ejemplos: 

Input : c1 = 5, c2 = 2, c3 = 3, c4 = 4
Output : 468
Explanation : We can form two 234s

Input : c1 = 5, c2 = 3, c3 = 1, c4 = 5
Output : 258
Explanation : We can form one 234 and two 12s

Enfoque : Un enfoque eficiente es intentar primero hacer 234’s. El número posible de 234s es mínimo de c2, c3, c4. Después de esto, con los 1 y 2 restantes, intente formar 12.

A continuación se muestra la implementación del enfoque anterior:  

// CPP program to maximum possible sum
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum possible sum
int Maxsum(int c1, int c2, int c3, int c4)
{
    // To store required sum
    int sum = 0;
 
    // Number of 234's can be formed
    int two34 = min(c2, min(c3, c4));
 
    // Sum obtained with 234s
    sum = two34 * 234;
 
    // Remaining 2's
    c2 -= two34;
 
    // Sum obtained with 12s
    sum += min(c2, c1) * 12;
 
    // Return the required sum
    return sum;
}
 
// Driver code
int main()
{
    int c1 = 5, c2 = 2, c3 = 3, c4 = 4;
 
    cout << Maxsum(c1, c2, c3, c4);
 
    return 0;
}

// Java program to maximum possible sum
class GFG
{
     
// Function to find the maximum possible sum
static int Maxsum(int c1, int c2, int c3, int c4)
{
    // To store required sum
    int sum = 0;
 
    // Number of 234's can be formed
    int two34 = Math.min(c2,Math.min(c3, c4));
 
    // Sum obtained with 234s
    sum = two34 * 234;
 
    // Remaining 2's
    c2 -= two34;
 
    // Sum obtained with 12s
    sum +=Math.min(c2, c1) * 12;
 
    // Return the required sum
    return sum;
}
 
// Driver code
public static void main(String[] args)
{
    int c1 = 5, c2 = 2, c3 = 3, c4 = 4;
 
    System.out.println(Maxsum(c1, c2, c3, c4));
}
}
 
// This code is contributed by Code_Mech.

# Python3 program to maximum possible sum
 
# Function to find the maximum
# possible sum
def Maxsum(c1, c2, c3, c4):
 
    # To store required sum
    sum = 0
 
    # Number of 234's can be formed
    two34 = min(c2, min(c3, c4))
 
    # Sum obtained with 234s
    sum = two34 * 234
 
    # Remaining 2's
    c2 -= two34
    sum += min(c2, c1) * 12
 
    # Return the required sum
    return sum
 
# Driver Code
c1 = 5; c2 = 2; c3 = 3; c4 = 4
print(Maxsum(c1, c2, c3, c4))
 
# This code is contributed by Shrikant13

// C# program to maximum possible sum
using System;
 
class GFG
{
     
// Function to find the maximum possible sum
static int Maxsum(int c1, int c2, int c3, int c4)
{
    // To store required sum
    int sum = 0;
 
    // Number of 234's can be formed
    int two34 = Math.Min(c2, Math.Min(c3, c4));
 
    // Sum obtained with 234s
    sum = two34 * 234;
 
    // Remaining 2's
    c2 -= two34;
 
    // Sum obtained with 12s
    sum +=Math.Min(c2, c1) * 12;
 
    // Return the required sum
    return sum;
}
 
// Driver code
public static void Main()
{
    int c1 = 5, c2 = 2, c3 = 3, c4 = 4;
 
    Console.WriteLine(Maxsum(c1, c2, c3, c4));
}
}
 
// This code is contributed
// by Akanksha Rai

<?php
// PHP program to maximum possible sum
 
// Function to find the maximum possible sum
function Maxsum($c1, $c2, $c3, $c4)
{
    // To store required sum
    $sum = 0;
 
    // Number of 234's can be formed
    $two34 = min($c2, min($c3, $c4));
 
    // Sum obtained with 234s
    $sum = $two34 * 234;
 
    // Remaining 2's
    $c2 -= $two34;
 
    // Sum obtained with 12s
    $sum += min($c2, $c1) * 12;
 
    // Return the required sum
    return $sum;
}
 
// Driver code
$c1 = 5; $c2 = 2;
$c3 = 3; $c4 = 4;
 
echo Maxsum($c1, $c2, $c3, $c4);
 
// This code is contributed by Ryuga
?>

<script>
// Java Script program to maximum possible sum
 
// Function to find the maximum possible sum
function Maxsum(c1,c2,c3,c4)
{
    // To store required sum
    let sum = 0;
 
    // Number of 234's can be formed
    let two34 = Math.min(c2,Math.min(c3, c4));
 
    // Sum obtained with 234s
    sum = two34 * 234;
 
    // Remaining 2's
    c2 -= two34;
 
    // Sum obtained with 12s
    sum +=Math.min(c2, c1) * 12;
 
    // Return the required sum
    return sum;
}
 
// Driver code
 
    let c1 = 5, c2 = 2, c3 = 3, c4 = 4;
 
    document.write(Maxsum(c1, c2, c3, c4));
 
// This code is contributed by sravan kumar G
</script>

468

Complejidad de tiempo: O(1)

Espacio Auxiliar: O(1)