I tried but none of my approach worked and some are giving TLE. Please help
Please tell me the approach
Hi @aakshatmalhotra100
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.
- Letβs talk about the singular case when we are considering the max window for only 'aβs and consider only the swapping of b-> a. If we are able to get the answer for max window of consecutive 'aβs , we can simply implement the same algo for the max βbβ window as well.
- So we started i from 0 and j from k.
- Move j ahead freely as long as there are βaβ characters at s[ j ] position.
- Maintain a count variable which counts the number of swaps made or the number of 'bβs in our A window.
- If you encounter a βbβ 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 βbβ as well.
- Output the maximum size window of βaβ and βbβ.
refer this -->
I hope Iβve cleared your doubt. I ask you to please rate your experience here
Your feedback is very important. It helps us improve our platform and hence provide you
the learning experience you deserve.
On the off chance, you still have some questions or not find the answers satisfactory, you may reopen
the doubt.