Your Name

The whole problem is asking whether $t$ is an anagram of $s$.

That is it. No hidden romance mechanic, no cursed swap restriction, no anime lore check. If we can rearrange the letters of $s$ however we want, then the only thing that matters is the multiset of characters.

Key Observation

A rearrangement can change positions, but it cannot change letter counts.

For example, from humitsa, we can rearrange letters into mitsuha because both strings contain:

• one h
• one u
• one m
• one i
• one t
• one s
• one a

Same letters, same counts, different order. Totally fine.

But aakima cannot become makima, because:

• aakima has three as and no m? Actually it has one m, one k, one i, and three as.
• makima has two ms, one a, one k, one i, one a total two as.

The counts do not match, so no amount of swapping can fix it. Rearranging is powerful, but it is not alchemy.

How To Check This

There are two standard ways:

1. Count frequencies of all 26 lowercase letters.
2. Sort both strings and compare them.

Since $n \le 20$, sorting is absurdly cheap. We can just do:

• sort $s$
• sort $t$
• if $s = t$, answer YES
• otherwise answer NO

Why does sorting work?

Sorting puts equal letters into a canonical order. Any two strings with the same letters and counts will become exactly the same sorted string.

For example:

misaka sorted becomes aaikms

mikasa sorted also becomes aaikms

So the answer is YES.

Complexity

For each test case, sorting two strings of length $n$ costs:

$O(n \log n)$

With $n \le 20$ and $q \le 1000$, this is basically free. The memory usage is $O(1)$ aside from the input strings.