SanketAndStrings

#1

can you give me some hint how to solve

#2

See this detailed description will clear your doubt
Implement the following approach:
Perform following for both the characters a and b individually,

  1. Take two pointers 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 the element at lth index is different.
    3.2.1. increment l.
    3.3. Compare max with count for maximum value.
    3.4. increment r.

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
Also attaching code for same algorithm to get an idea.

#3

sir in your code you have implemented some other logic

#4

By mistake attached wrong code, see this

#5

sir the previous code is also correct only space complexity is used there.Can you please explain that approach also

#6

It’s an intuitive approach. Just try to do dry run for different test cases. You will get the intuition. One of the dry run is done here. You can see, it will help you in understanding it better.

#7

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