Related Articles

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

• Difficulty Level : Easy
• Last Updated : 06 Jul, 2021

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++

 `// CPP program to maximum possible sum` `#include ``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

 `// 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

 `# 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#

 `// 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

 ``

## Javascript

 ``
Output:
`468`

My Personal Notes arrow_drop_up