Please go through my code and let me know what in my approach is wrong. https://ide.codingblocks.com/s/363497
Also, a hint if possible!
Sanket and string!
Hey @yashjain0112
Implement the following approach:
Perform following for both the characters a and b individually,
- Take two pointers l and r to mark the left and right index of the string under consideration.
- starting from l=0,r=0,max=0,count=0.
- 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.
- 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.
- 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
1 Like
Here’s the code in which there are lot’s of comment for better understanding. You can refer this 
1 Like
The code itself is giving 2 for the example you gave
My code is not working, as i said!
Need help
The solution encoder gave, I tried coding it but didn’t work.
Used the code he gave, that gave wrong answer for k=1, s=abba
u can add one more case in the code to check whether the given string is already perfect or not.
if it is then simply lenght of string will be the required answer
Great support must say
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