What else I have to add in it

#1
#2

Hey @ashish_arora369
You approach is not correct.
say a>b
then
assume string is like this bbbbbbbaaaaaa for some k=2
now check if your code will correct answer or not.

Follow this approach:
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’.
#3

bbbbbbbaaaaaa for some k=2 it gave answer 9 according to me which is right

#4

No @ashish_arora369
Answer is 12 for that
We need to return max string length with equal chars
See last 12 char string
They have same no of both chars and we dont have to do any swap here

#5

may you show me how you got that answer

#6

Here is the correct algo , please refer to this :slight_smile:

#7

Hey @ashish_arora369
If this is resolved then please mark it so:)

#8

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