Error in this code

#1

#include
#include
#include
using namespace std;
void perfectness(char arr[], int k){
int fre[2]={0};
int maxfre = 0,maxlen=0, Start = 0, End = 0;
//char perfChar;
while(End<strlen(arr)){
if(arr[End]==‘a’){
fre[0]++;
}else if(arr[End]==‘b’){
fre[1]++;
}
maxlen=max(maxlen,End-Start+1);
maxfre=max(maxfre,fre[arr[End]-‘a’]);
if(End-Start+1-maxfre>k){
fre[arr[End]-‘a’]–;
Start++;
}
End++;
}
cout<<maxlen;
}
int main(){
int k;
cin >> k;
char arr[1000];
cin >> arr;
perfectness(arr,k);

return 0;

}

#2

hi @umnah1103_2205d5eee7973385
You can solve this problem in O(n) time using the two pointer approach.

  • Make two variabes , say i and j .
  • i defines the beginning of a window and j defines its end.
  • Start i from 0 and j from k.
  • Let’s talk about the singular case when we are considering the max window for only 'a’s and consider only the swapping of b-> a. If we are able to get the answer for max window of consecutive 'a’s , we can simply implement the same algo for the max ‘b’ window as well.
  • So we started i from 0 and j from k.
  • Move j ahead freely as long as there are ‘a’ characters at s[ j ] position.
  • Maintain a count variable which counts the number of swaps made or the number of 'b’s in our A window.
  • If you encounter a ‘b’ char at s[ j ] position , increment the count variable. Count should never exceed k .
  • Take the size of the window at every point using length = j - i + 1;
  • Compute the max size window this way and do the same for ‘b’ as well.
  • Output the maximum size window of ‘a’ and ‘b’.

refer this code -->

#3

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