HackerEarth A smallest number problem solution

In this HackerEarth A smallest number problem solution You are given an integer K.

Find the smallest number N such that  has exactly K digits and none of the digits in N is $$0$$. Also, the product of digits in number N is greater than or equal to the sum of digits in number N.

HackerEarth A smallest number problem solution.

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

int get_digits(int num)
{
    int answer = 0;
    while(num)
    {
        num /= 10;
        answer++;
    }
    return answer;
}

bool check(int num)
{
    sum = 0;
    product = 1;
    while(num)
    {
        sum += (num%10);
        product *= (num%10);
        num /= 10;
    }

    return (product >= sum);
}

void solve(){
    int k;
    cin >> k;
    assert(1 <= k and k <= 500000);

    if(k <= 6)
    {
        for(int i = 1 ; i < 1000000 ; i++){
            if(check(i) and get_digits(i) == k)
            {
                cout << i << endl;
                return;
            }
        }
    }
    else{
        for(int i = 1 ; i < 1000000 ; i++){
            check(i);
            if(product >= sum + k - get_digits(i))
            {
                for(int j = 1 ; j <= (k - get_digits(i)) ; j++){
                    cout << 1;
                }
                cout << i << endl;
                return;
            }
        }
    }
}
signed main(){
    int t;
    cin >> t;
    assert(1 <= t and t <= 10);
    while(t--){
        solve();
    }
}

Second solution

def multiplyList(myList):
    # Multiply elements one by one
    result = 1
    for x in myList:
        result = result * x
    return result


t = int(input())
while t > 0:
    t -= 1
    n = int(input())
    for i in range(1, 10 ** 6):
        if '0' not in str(i) and multiplyList(map(int, str(i))) >= n - len(str(i)) + sum(map(int, str(i))):
            print('1' * (n - len(str(i))), i, sep='')
            break