Cipher both approaches O(n) and O(n^2)

Jack and Daniel are friends.
They want to encrypt their conversation so that they can save themselves from interception by a detective agency. So they invent a new cipher.
Every message is encoded to its binary representation of length .
Then it is written down times, shifted by bits.
If and it looks so:

1001010   
 1001010  
  1001010 
   1001010

Then calculate XOR in every column and write it down. This number is called . For example, XOR-ing the numbers in the above example results in

1110100110

Then the encoded message and are sent to Daniel.
Jack is using this encoding algorithm and asks Daniel to implement a decoding algorithm.
Can you help Daniel implement this?
Input Format
The first line contains two integers and .
The second line contains string of length consisting of ones and zeros.
Output Format
Decoded message of length , consisting of ones and zeros.
Constraints



It is guaranteed that is correct.
Sample Input#00

7 4
1110100110

Sample Output#00

1001010

Sample Input#01

6 2
1110001

Sample Output#01

101111

Explanation
Input#00

1001010
 1001010
  1001010
   1001010
----------
1110100110

Input#01

101111
 101111
-------
1110001

this is O(n²) approach

)

#include <stdio.h>

#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
    int  n , k ;
    scanf("%d %d",&n,&k);
   
    char a[n+k];
    scanf("%s",a);
    a[n]='\0';
    char final[n+1];

    final[0]= a[0];
    final[n]= a[n];
  
    printf("%c",final[0]);
    for (int i = 1;i<n;i++)
    {
        final[i]='0';
        int answer = 0;
        for (int m = i-1;((m>=0)&&(m>=(i-k+1)));m--)       
        {answer = (final[m]-'0') ^ answer  ;
        }
        final[i] = ( (a[i]-'0') ^ answer ) +'0';
        final[i]
        printf("%c",final[i]);
    }
   
    return 0;
}

approach (2)  this is O(N) 

 #include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
    int  n , k ;
    scanf("%d %d",&n,&k);
   
    char a[n+k];
    scanf("%s",a);
    a[n]='\0';
    char final[n+1];

    final[0]= a[0];
    final[n]= a[n];
  
    printf("%c",final[0]);
    int answer = 0;
    for (int i = 1;i<n;i++)
    {
        final[i]='0';       
       answer = (final[i-1]-'0') ^ answer  ;
        if (i>=k)
        {
            answer = answer ^(final[i-k]-'0');
        }
        final[i] = ( (a[i]-'0') ^ answer ) +'0';
        printf("%c",final[i]);
    }
   
    return 0;
}