Reorganizar una string en forma de suma entera seguida del carácter minimizado

Dada una string que incluye letras en minúsculas y dígitos numéricos. La tarea es construir otra string que consista en la suma de dígitos seguida de la suma de todos los alfabetos minimizados a un solo carácter. Si no hay ningún dígito numérico presente, agregue 0 a la string. 
Nota: La suma alfabética se realiza de esta manera: a+a = b, d+y = c. 
Ejemplos: 
 

Input: str = "ab37b3a8"
Output: 21f
Sum of digits = 3 + 7 + 3 + 8 = 21
Sum of alphabets = a + b + b + a = 1 + 2 + 2 + 1 = 6
Alphabet at 6th position is f.

Input: str = "3by2b2a2"
Output: str = 9d

Enfoque: para separar dígitos y alfabetos, recorra la string dada, si el carácter es numérico, agréguelo a digit-sum y si es un alfabeto, agréguelo a alphabet-sum. 
 

1. Comience a atravesar la cuerda.
2. Si el dígito es numérico, agregue su valor (str[i] – ‘0’) a digitSum.
3. De lo contrario, agregue str[i]-‘a’+1 a alphabetSum.
4. Convierta alphabetSum a char y agréguelo al formato de caracteres de digitSum .

A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// function to return maximum volume
string separateChar(string str)
{
    int n = str.size(), digitSum = 0;
    int alphabetSum = 0, j = 0;
 
    // separate digits and alphabets
    for (int i = 0; i < n; i++) {
        if (isdigit(str[i]))
            digitSum += str[i] - '0';
 
        else {
            alphabetSum += str[i] - 'a' + 1;
            alphabetSum %= 26;
        }
    }
 
    // change digit sum to string
    string sumStr = to_string(digitSum);
 
    // change alphabet sum to string
    char alphabetStr = char(alphabetSum + 'a' - 1);
 
    // concatenate sum to alphabets string
    sumStr += alphabetStr;
 
    return sumStr;
}
 
// Driver code
int main()
{
    string str = "3652adyz3423";
    cout << separateChar(str);
    return 0;
}

// Java implementation of above approach
import java.util.*;
class Solution
{
 
// function to return maximum volume
static String separateChar(String str)
{
    int n = str.length(), digitSum = 0;
    int alphabetSum = 0, j = 0;
 
    // separate digits and alphabets
    for (int i = 0; i < n; i++) {
        if (str.charAt(i)>='0'&&str.charAt(i)<='9') {
            digitSum += (int)(str.charAt(i) - '0');
        }
        else {
            alphabetSum += str.charAt(i) - 'a' + 1;
            alphabetSum %= 26;
        }
    }
 
    // change digit sum to string
    String sumStr = ""+(digitSum);
 
    // change alphabet sum to string
    char alphabetStr = (char)(alphabetSum + 'a' - 1);
 
    // concatenate sum to alphabets string
    sumStr += alphabetStr;
 
    return sumStr;
}
 
// Driver code
public static void main(String args[])
{
    String str = "3652adyz3423";
    System.out.println(separateChar(str));
     
}
 
}
//contributed by Arnab Kundu

# Python 3 implementation of the above approach
 
# function to return maximum volume
def separateChar(str__):
    n = len(str__)
    digitSum = 0
    alphabetSum = 0
    j = 0
 
    # separate digits and alphabets
    for i in range(n):
        if (ord(str__[i]) >= 48 and
            ord(str__[i]) <= 56):
            digitSum += ord(str__[i]) - ord('0')
 
        else:
            alphabetSum += ord(str__[i]) - ord('a') + 1
            alphabetSum %= 26
     
    # change digit sum to string
    sumStr = str(digitSum)
 
    # change alphabet sum to string
    alphabetStr = chr(alphabetSum + ord('a') - 1)
 
    # concatenate sum to alphabets string
    sumStr += alphabetStr
 
    return sumStr
 
# Driver code
if __name__ == '__main__':
    str__ = "3652adyz3423"
    print(separateChar(str__))
     
# This code is contributed by
# Shashank_Sharma

// C# implementation of above approach
using System;
public class Solution{
     
    // function to return maximum volume
    static String separateChar(String str)
    {
        int n = str.Length, digitSum = 0;
        int alphabetSum = 0, j = 0;
 
        // separate digits and alphabets
        for (int i = 0; i < n; i++) {
            if (str[i]>='0'&&str[i]<='9') {
                digitSum += (int)(str[i] - '0');
            }
            else {
                alphabetSum += str[i] - 'a' + 1;
                alphabetSum %= 26;
            }
        }
 
        // change digit sum to string
        String sumStr = ""+(digitSum);
 
        // change alphabet sum to string
        char alphabetStr = (char)(alphabetSum + 'a' - 1);
 
        // concatenate sum to alphabets string
        sumStr += alphabetStr;
 
        return sumStr;
    }
 
    // Driver code
    public static void Main()
    {
        String str = "3652adyz3423";
        Console.WriteLine(separateChar(str));
 
    }
}
 
// This code is contributed by 29AjayKumar

<script>
 
// Javascript implementation of the above approach
 
// function to return maximum volume
function separateChar(str)
{
    var n = str.length, digitSum = 0;
    var alphabetSum = 0, j = 0;
 
    // separate digits and alphabets
    for (var i = 0; i < n; i++) {
        if (str[i] >= '0' && str[i] <= '9')
            digitSum += (str[i].charCodeAt(0) - '0'.charCodeAt(0));
 
        else {
            alphabetSum += (str[i].charCodeAt(0) - 'a'.charCodeAt(0) + 1);
            alphabetSum %= 26;
        }
    }
 
    // change digit sum to string
    var sumStr = digitSum.toString();
 
    // change alphabet sum to string
    var alphabetStr = String.fromCharCode(alphabetSum + 'a'.charCodeAt(0) - 1);
 
    // concatenate sum to alphabets string
    sumStr += alphabetStr;
 
    return sumStr;
}
 
// Driver code
var str = "3652adyz3423";
document.write( separateChar(str));
 
</script>

28d