Have you tried to investigate how sumOfDigits(n) % 10 behaves when you iterate n = 1..100 or something like this? You probably can come upon the idea...

maybe here is a simple way :

long long findNthNumber(int n) {
    long long num = 0, cnt = 0;
    while (cnt < n) {
        num++;
        if (sumOfDigits(num) % 10 == 0) {
            cnt++;
        }
    }
    return num;
}

I think it can be done in O(1): If k is the sum of digits of n, then the final answer will be concatenation of n and digit q = (10 — k % 10) % 10

Another gray trying to solve dp before knowing how to sort array :((