Sanket and string .. although i have found dp states but how will i do in given time limits

#1

size od dp table gets out of limit when i m cosiderring whole range of input … please help me sir…

#2

@codingblockscpp
This is a simple two pointer approach problem that can be solved in O(n) time using two pointers. No need to use DP for it , especially 4D - DP.

#3

thank you sir … i am not getting approach … can you help me please

#4

@codingblockscpp
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’.

#5

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