2578. Split With Minimum Sum

Given a positive integer num, split it into two non-negative integers num1 and num2 such that:

• The concatenation of num1 and num2 is a permutation of num.

• In other words, the sum of the number of occurrences of each digit in num1 and num2 is equal to the number of occurrences of that digit in num.

• num1 and num2 can contain leading zeros.

Return the minimum possible sum of num1 and num2.

Notes:

• It is guaranteed that num does not contain any leading zeros.
• The order of occurrence of the digits in num1 and num2 may differ from the order of occurrence of num.

class Solution {
public:
    int splitNum(int num) {
        long long res = 0, po = 1;
        string s = to_string(num);
        sort(begin(s), end(s));
        while(s.length()) {
            if(s.length() == 1) {
                res += (s.back() - '0') * po;
                s.pop_back();
            } else {
                res += (s.back() - '0') * po;
                s.pop_back();
                res += (s.back() - '0') * po;
                s.pop_back();
            }
            po *= 10;
        }
        return res;
    }
};