Kartik bhaiya and strings

#1

Please tell approach of the problem

#2

@kumarakash121005
you need to find the maximum length of a substring (consecutive subsequence) consisting of equal letters. and u are allowed to change no more than k characters of the original string.

for example->
4 2
abba

answer is 4 becuase can obtain both strings “aaaa”(by change 2 b’s into a) and “bbbb”(changing a’ to b)

8 1
aabaabaa
answer is 5 becuase we can obtain string “aaaaabaa” or string “aabaaaaa”. (both have 5 length substring with all smae character)

You can solve this problem using two pointer technique:
Let the first pointer is l and the second pointer is r . Then for every position l we will move right end r until on the substring s l s l+ 1 … s r it is possible to make no more than k swaps to make this substring beautiful. Then we need to update the answer with length of this substring and move l to the right.

  • 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

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