# Minimum Absolute Score Codechef Solution

### We Are Discuss About CODECHEF 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.

### Constraints

• 1 \leq T \leq 100
• 1 \leq N \leq 10^5
• Both strings A and B have same length N and contain only lowercase English letters.
• The sum of N over all test cases won’t exceed 10^5.

### Sample 1:

Input

Output

4
3
abb
baz
3
zzc
aaa
4
fxbs
dkrc
5
eaufq
drtkn

2
0
11
9


### 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 1z\to a for a cost of +1
• Operation 1 at index 2z\to a for a cost of +1
• Operation 2 at index 3a\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.

## Minimum Absolute Score Codechef Solution

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