Unable to solve

#1

please provide sum hint

#2

@kartik-malik
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

why are we starting j from k and i from 0

#4

@kartik-malik because i is start of the window and j end (2nd point)

#5

can u please explain through a example

#6

also what will happen to characters beteen i and j

#7

@kartik-malik its diff to explain(i mean dry run here) if you are having difficulty understanding it refer to the editorial code ( dry run it for some example) if any doubt in it ask


(its well commented every step)

#8

does this question comes in easy category :sweat:

#9

@kartik-malik yes i understand its a little tough to understand :sweat_smile:
but once done it will become clear. you will have to do it again after some time without looking for better grasp

#10

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