## GUPTA MECHANICAL

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# [Solution] Minimum Absolute Score CodeChef Solution

## Problem

You are given two strings $A$ and $B$ of length $N$ consisting of lowercase English letters. Your objective is to make both the strings equal.

You can apply one of the following $2$ operations at each index $i$:

• Convert char $A_i$ to $B_i$ by doing right cyclic shift of character $A_i$. This increases your score by amount equal to cyclic shifts done.
• Convert char $B_i$ to $A_i$ by doing right cyclic shift of character $B_i$. This decreases your score by amount equal to cyclic shifts done.

If the operations are applied optimally, find the minimum absolute score possible after making both the strings equal.

Note: A single right cyclic shift converts one character to the next in alphabetical order, except for $z$ which goes to $a$. That is, the sequence looks like

$a \to b \to c \to \ldots \to y \to z \to a \to b \to \ldots$

So, for example converting $a$ to $e$ requires $4$ right cyclic shifts, and converting $k$ to $i$ requires $24$.

### Input Format

• The first line of input will contain a single integer $T$, denoting the number of test cases.
• Each test case consists of three lines of input.
• The first line of each test case contains one integer $N$ — the length of strings $A$ and $B$.
• The second line contains string $A$.
• The third line contains string $B$.

### Output Format

For each test case, output on a new line the minimum absolute score possible after making both the strings equal.

### Explanation:

Test case $1$: The minimum absolute score can be obtained as follows:

• Apply operation $1$ at position $1$, converting $a$ to $b$ for a cost of $+1$.
• Apply operation $2$ at position $2$, converting $a$ to $b$ for a cost of $-1$.
• Apply operation $2$ at position $3$, converting $z$ to $b$ for a cost of $-2$.

The score is then $1 -1 -2 = -2$, with absolute value $2$. This is the lowest possible absolute value attainable.

Test case $2$: Apply operations as follows:

• Operation $1$ at index $1$$z\to a$ for a cost of $+1$
• Operation $1$ at index $2$$z\to a$ for a cost of $+1$
• Operation $2$ at index $3$$a\to c$ for a cost of $-2$

This gives us a final score of $1 + 1 - 2 = 0$, which has absolute value $0$. It is not possible to do better than this.