Equal Frequencies
Contest: VK Cup 2022 - Отборочный раунд (Engine)
Difficulty rating: 1600
Tags: brute force, constructive algorithms, greedy, implementation, sortings, strings

Let's call a string balanced if all characters that are present in it appear the same number of times. For example, "coder", "appall", and "ttttttt" are balanced, while "wowwow" and "codeforces" are not.

You are given a string $$$s$$$ of length $$$n$$$ consisting of lowercase English letters. Find a balanced string $$$t$$$ of the same length $$$n$$$ consisting of lowercase English letters that is different from the string $$$s$$$ in as few positions as possible. In other words, the number of indices $$$i$$$ such that $$$s_i \ne t_i$$$ should be as small as possible.

Input format:
Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). The description of the test cases follows.

Each test case consists of two lines. The first line contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of the string $$$s$$$.

The second line contains the string $$$s$$$ of length $$$n$$$ consisting of lowercase English letters.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.

Output format:
For each test case, print the smallest number of positions where string $$$s$$$ and a balanced string $$$t$$$ can differ, followed by such a string $$$t$$$.

If there are multiple solutions, print any. It can be shown that at least one balanced string always exists.

Constraints:
Time limit: 2.0 s
Memory limit: 512.0 MB

Note:
In the first test case, the given string "hello" is not balanced: letters 'h', 'e', and 'o' appear in it once, while letter 'l' appears twice. On the other hand, string "helno" is balanced: five distinct letters are present in it, and each of them appears exactly once. Strings "hello" and "helno" differ in just one position: the fourth character. Other solutions are possible too.

In the second test case, string "codefofced" is balanced since only letters 'c', 'o', 'd', 'e', and 'f' are present in it, and each of them appears exactly twice.

In the third test case, string "eeeee" is balanced since only letter 'e' is present in it.

In the fourth test case, the given string "appall" is already balanced.

Example input:
4
5
hello
10
codeforces
5
eevee
6
appall

Example output:
1
helno
2
codefofced
1
eeeee
0
appall

Editorial:
Instead of "finding $$$t$$$ that differs from $$$s$$$ in as few positions as possible", let's formulate it as "finding $$$t$$$ that matches $$$s$$$ in as many positions as possible", which is obviously the same.
First of all, let's fix $$$k$$$, the number of distinct characters string $$$t$$$ will have. Since the string must consist of lowercase English letters, we have $$$1 \le k \le 26$$$, and since the string must be balanced, we have $$$n \bmod k = 0$$$. For each $$$k$$$ that satisfies these conditions, we will construct a balanced string that matches $$$s$$$ in as many positions as possible. In the end, out of all strings we will have constructed, we will print the one with the maximum number of matches.
From now on, we are assuming $$$k$$$ is fixed. Suppose we choose some character $$$c$$$ to be present in string $$$t$$$. We need to choose exactly $$$\frac{n}{k}$$$ positions in $$$t$$$ to put character $$$c$$$.
Let $$$\operatorname{freq}_{c}$$$ be the number of occurrences of $$$c$$$ in $$$s$$$.
Then, in how many positions can we make $$$s$$$ and $$$t$$$ match using character $$$c$$$? The answer is: in $$$\min(\frac{n}{k}, \operatorname{freq}_{c})$$$ positions.
Now, since we want to maximize the total number of matches, we should choose $$$k$$$ characters with the largest values of $$$\min(\frac{n}{k}, \operatorname{freq}_{c})$$$. This is also equivalent to choosing $$$k$$$ characters with the largest values of $$$\operatorname{freq}_{c}$$$.
How to construct the desired string? For each chosen character $$$c$$$, pick any $$$\min(\frac{n}{k}, \operatorname{freq}_{c})$$$ of its occurrences in $$$s$$$ and put $$$c$$$ in the corresponding positions in $$$t$$$. Then, if $$$\operatorname{freq}_{c} < \frac{n}{k}$$$, save the information about $$$\frac{n}{k} - \operatorname{freq}_{c}$$$ unused characters $$$c$$$; otherwise, if $$$\operatorname{freq}_{c} > \frac{n}{k}$$$, save the information about $$$\operatorname{freq}_{c} - \frac{n}{k}$$$ empty positions in $$$t$$$. In the end, match the unused characters with the empty positions arbitrarily.