Code Of The Day: Google Kick Start Problem

 

K-Goodness String

Problem

Charles defines the goodness score of a string as the number of indices i$i$ such that Si≠SN−i+1${\mathbf{S}}_i\ne {\mathbf{S}}_{\mathbf{N}-i+1}$ where 1≤i≤N/2$1\le i\le \mathbf{N}/2$ (1$1$-indexed). For example, the string CABABC has a goodness score of 2$2$ since S2≠S5${\mathbf{S}}_2\ne {\mathbf{S}}_5$ and S3≠S4${\mathbf{S}}_3\ne {\mathbf{S}}_4$.

Charles gave Ada a string S$S$ of length N$N$, consisting of uppercase letters and asked her to convert it into a string with a goodness score of K$K$. In one operation, Ada can change any character in the string to any uppercase letter. Could you help Ada find the minimum number of operations required to transform the given string into a string with goodness score equal to K$K$?

Input

The first line of the input gives the number of test cases, T$T$. T$T$ test cases follow.

The first line of each test case contains two integers N$N$ and K$K$. The second line of each test case contains a string S$S$ of length N$N$, consisting of uppercase letters.

Output

For each test case, output one line containing Case #x$x$: y$y$, where x$x$ is the test case number (starting from 1) and y$y$ is the minimum number of operations required to transform the given string S$S$ into a string with goodness score equal to K$K$.

Limits

Memory limit: 1 GB.
1≤T≤100$1\le \mathbf{T}\le 100$.
0≤K≤N/2$0\le \mathbf{K}\le \mathbf{N}/2$.

Test Set 1

Time limit: 20 seconds.
1≤N≤100$1\le \mathbf{N}\le 100$.

Test Set 2

Time limit: 40 seconds.
1≤N≤2×105$1\le \mathbf{N}\le 2\times {10}^5$ for at most 10$10$ test cases.
For the remaining cases, 1≤N≤100$1\le \mathbf{N}\le 100$.

Sample

Sample Input

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2
5 1
ABCAA
4 2
ABAA

Sample Output

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Case #1: 0
Case #2: 1

In Sample Case #1, the given string already has a goodness score of 1$1$. Therefore the minimum number of operations required is 0$0$.

In Sample Case #2, one option is to change the character at index 1$1$ to B in order to have a goodness score of 2$2$. Therefore, the minimum number of operations required is 1$1$.

n = int(input())

l =[]

s = []

for i in range(n):

    x,y  =  input().split()

    l.append([ int(x) , int(y)])

    s.append(input())

def main(s,m,k):

    k = int(k)

    minop= 0

    x = 0

    for i in range(m//2):

        if s[i] != s[m-i-1]:

            x+=1 

    if x==k:

        return 0 

    else:

        return abs(x-k)

for i in range( 0,len(s)):

    mk,nk = l[i]

    j = s[i]

    print(f"Case #{i+1}: {main(j,mk,nk)}")