Algorithm:

1. Declare sum=0 to add integers of the string.
2. Declare map<char,int>m to tore characters of the string with respective frequencies;

3. For i=0:s.length()-1
       IF(s[i]>='0' && s[i]<='9') //current character is digit 
           sum+=s[i]-'0'; //add to sum//s[i]-'0'=actual numeric value
       Else //it's a alphabetical character
           m[s[i]]++; //store character with its frequency
       END IF
   END FOR
   

4. Rearranging part:

   For iterator it=m.begin():m.end()-1
       //inner loop to print the character as many times it occurs
       For i=0 : it->second-1 //it->second is the frequency of character it->first
           Print it->first;
       END FOR
   END FOR
   Print sum;
   

Example with explanation:

    Input:
    XY3BFA9KA2

    So,
    After processing :

    Map m becomes:
    Char(key)	Int(value)
    'A'	        2
    'B'	        1
    'F'	        1
    'K'	        1
    'X'	        1
    'Y'	        1

    Sum: 14

    Thus it prints:
    AABFKXY14

Note: The map is always sorted according to its key. Thus, here we don't need to take any additional care for maintain alphabet order as the map itself is always sorted as per its key (character here).

C++ implementation:

#include <bits/stdc++.h>
using namespace std;

void rearrange(string s){
	int sum=0;
	map<char,int> m;
	
	for(int i=0;i<s.length();i++){
		if(s[i]>='0' && s[i]<='9')//digut
			sum+=s[i]-'0';
		else //character
			m[s[i]]++;
	}

	for(auto it=m.begin();it!=m.end();it++){
		//print the character as many time it occured
		for(int i=0;i<it->second;i++)
			cout<<it->first;
	}
	cout<<sum<<endl; //printing sum
}

int main(){
	string s;

	cout<<"Enter input string\n";
	cin>>s;
	cout<<"Rearranged string is:\n";
	rearrange(s);

	return 0;
}

Output

Enter input string
XY3BFA9KA2
Rearranged string is:
AABFKXY14