HackerEarth Divide the digits problem solution

In this HackerEarth Divide, the digits problem solution You are given a number N. You are required to form two numbers X and Y such that:

The sum of frequency of each digit in X and Y is equal to frequency of that digit in N.
The sum of numbers X and Y must be minimum.

Your task is to determine the minimum possible sum of X and Y.

HackerEarth Divide the digits problem solution.

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

void solve(){
    int n;
    cin >> n;
    assert(10 <= n and n <= 2000000000000000000);

    vector<int>dig;
    while(n)
    {
        dig.push_back(n % 10);
        n /= 10;
    }

    sort(dig.begin(), dig.end());
    int sz = dig.size();
    int first = 0;
    int second = 0;
    for(int i = 0 ; i < sz ; i++)
    {
        if(i % 2)
        {
            first = (first*10 + dig[i]);
        }
        else{
            second = (second*10 + dig[i]);
        }
    }

    cout << (first + second) << endl;
}

signed main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    assert(1 <= t and t <= 100000);
    while(t--){
        solve();
    }
}

Second solution

t = int(input())
while t > 0:
    t -= 1
    n = list(input())
    n.sort()
    print(int(''.join(n[::2])) + int(''.join(n[1::2])))

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