Please tell me where im getting wrong all the test cases according me are correct and please tell me variation in this code not a new concept please

#1

Please tell me where im getting wrong all the test cases according me are correct and with deep respect please tell me variation in this code not a new concept please.

#2

hey @pssharma1410
debug your code for
1
abba
Output:
3
you get 4.
hope this helps :slight_smile:

#3

mam by my code im getting 3 not 4

#4
#5

my bad,for test case
2
abbbababbaba

answer should be 8 not 9

similarly for
2
aabbaababba
ans should be 6
debug for this and let me know if this doesn’t help

#6

for test case 2 abbbababbaba

#7

the maximum of a or b is 7 because b is occurring 7 times and since k is equal to 2 then we can replace to ‘a’ to get 9 'b’finally how the answer can be 8?? please explain

#8

look at the input carefully

  • abbbababbaba is the input
  • abbbbbbbbaba is the reason for 8

yes there are 7 b and u can replace 2 a but the only way you can replace them to get a maximum length substring(continuous) of equal characters gives us 8

#9

Mam now please tell me where im getting wrong https://ide.codingblocks.com/s/464485

#10

The code is complicated but the outputs should be correct please check it

#11

dry run your code for this test case:
1
abbaa
correct ouput should be 2
your code is not ideally correct:

as you have tried a lot you can read the correct way to solve the question and move forward and remember it

Approach - Two pointer approach

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’.