Bhaiya, is question mai thoda hint provide kar dijiye.
Kartik bhaiya and string
hi @ritikbunty2511_729d0258992e0982,
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 -->
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https://ide.codingblocks.com/s/681734 bhaiya is code ka bhi time complexity o(n) he hai, to ye code bhi sahi hai na ???
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