Sanket and strings

#1

i am unable to find the approach to solve it…

#2

Hello @ashutoshu,

You can implement the following approach:
Perform following for both the characters a and b individually,

  1. Take two pointer l and r to mark the left and right index of the string under consideration.
  2. starting from l=0,r=0,max=0,count=0.
  3. repeat until r <n(length of the string)
    3.1. increase the count whenever you find a different character(by different we mean if we are forming a string of a only, then b is different).
    3.2. while count is greater than k,
    3.2.1. decrement the count by one if element at lth index is different.
    3.2.1. increment l.
    3.3. Compare max with count for maximum value.
    3.4. increment r.

Hope, this would help.
Give a like, if you are satisfied.

#3

i am not understanding the logic

#4

Okay, @ashutoshu

Let’s understand this with an example:
Example:
1
abba
Output:
3
Explanation:

l=0: variable to mark the left index and initialized to 0
r=0: variable to mark the right index and initialized to 0
m=0: to store the perfectness and initialised to zero
t=k: to keep track of the number of replacements we can perform at any instance of time.

  1. First, we will check for the string of character ‘a’:

    character at a[r] is ‘a’, increment r
    r:1
    the character at a[r] is not ‘a’ and t>0, increment r and decrement t as we can replace one ‘b’ to ‘a’
    t:0 r:2
    the character at a[r] is not ‘a’ and t==0. So, we cannot replace. Thus replace m by max. of m and r-l.
    Repeat, while t==0:
    A. if a[l]!=‘a’: increment t (because we must have done one replacement for this before and now as we are not considering this character so we can use this replacement somewhere else.)
    B. increment l for each iteration of the loop.
    m:2 t:0 l:1 t:1 l:2
    the character at a[r] is not ‘a’ and t>0, increment r and decrement t as we can replace one ‘b’ to ‘a’
    t:0 r:3
    character at a[r] is ‘a’, increment r
    r:4
    replace m by max. of m and r-l.
    m:2

l=0: variable to mark the left index and initialized to 0
r=0: variable to mark the right index and initialized to 0
m=0: to store the perfectness and initialised to zero
t=k: to keep track of the number of replacements we can perform at any instance of time.

  1. Now, we will check for the string of character ‘b’:

    the character at a[r] is not ‘b’ and t>0, increment r and decrement t as we can replace one ‘a’ to ‘b’
    **t:0 r:1 **
    character at a[r] is ‘b’, increment r
    **r:2 **
    character at a[r] is ‘b’, increment r
    **r:3 **
    the character at a[r] is not ‘b’ and t==0. So, we cannot replace. Thus replace m by max. of m and r-l.
    Repeat, while t==0:
    A. if a[l]!=‘b’: increment t (because we must have done one replacement for this before and now as we are not considering this character so we can use this replacement somewhere else.)
    B. increment l for each iteration of the loop.
    **m:3 t:1 l:1 **
    the character at a[r] is not ‘b’ and t>0, increment r and decrement t as we can replace one ‘a’ to ‘b’
    **t:0 r:4 **
    replace m by max. of m and r-l.
    m:3

Hence, the output is m:
3

Hope, this would help.
Give a like if you are satisfied.

#5

output of this program is 4 not 3.and in the 1st case ,how t becomes 0 from 2 1st time??

#6

No, @ashutoshu

The output is 3 only.
Please observe the example i have explained.
It is different from the one given in question.
The value of k is 1 instead of 2.

Hope, this would help.
Give a like if you are satisfied.

#7

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