Only One Digit

We need the smallest non-negative integer $y$ such that $x$ and $y$ share at least one digit.

The important thing is to not overcomplicate this. Since $y$ can be any non-negative integer, it can be a single digit: $0,1,2,\dots,9$.

If a digit $d$ appears in $x$, then choosing $y=d$ immediately works, because $y$'s representation contains exactly that digit $d$.

For example:

• $x=96$: digits are $9$ and $6$, so $y=6$ works, and it is smaller than $9$.
• $x=122$: digits are $1$ and $2$, so $y=1$ works.
• $x=1000$: digits are $1$ and $0$, and because $y$ may be non-negative, $y=0$ works. This is the classic tiny trap. Positive-only brain gets cooked here.

Key observation: the answer is always one digit.

Why? Because $x$ has at least one digit. Pick any digit from $x$, call it $d$. Then $y=d$ is valid and $0 \le d \le 9$. So there is always a valid answer at most $9$.

Any number with two or more digits is at least $10$, so it can never beat a valid one-digit answer. Therefore, we only care about digits already inside $x$.

So the smallest valid $y$ is simply:

$$\min(\text{digits of } x)$$

Implementation is straightforward:

1. Read $x$.
2. Initialize ans to something large, like 9 or 10.
3. Repeatedly take the last digit using x % 10.
4. Update ans with the minimum digit seen.
5. Remove the last digit using x /= 10.
6. Print ans.

Because $1 \le x \le 1000$, this is basically instant. The complexity per test case is $O(\text{number of digits in }x)$, which is at most $4$ here. So yeah, the computer is not even waking up for this one.