Given count of digits 1, 2, 3, 4, find the maximum sum possible

Given the count of digits 1, 2, 3, 4. Using these digits you are allowed to only form numbers 234 and 12. The task is to find the maximum possible sum that can be obtained after forming the numbers.

Note: The aim is only to maximize the sum, even if some of the digits left unused.

Examples:

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


Approach : An efficient approach is to first try to make 234’s. The possible number of 234s are minimum of c2, c3, c4. After this, with remaining 1’s and 2’s try to form 12s.

Below is the implementation of the above approach :

C++

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// 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 requied sum
    return sum;
}
  
// Driver code
int main()
{
    int c1 = 5, c2 = 2, c3 = 3, c4 = 4;
  
    cout << Maxsum(c1, c2, c3, c4);
  
    return 0;
}

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Java

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// 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 requied 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.

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Python3

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# 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 requied 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

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C#

// 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 requied 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

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<?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 requied 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
?>

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Output:

468


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