Sanket and strings

#1

I CAN’T UNDERSTAND THIS QUESTION. PLZ EXPLAIN WHAT WE HAVE TO DO HERE

#2

Hello @sharma.shubham,

Explanation of the question:

  1. You have given a string and a number k.
  2. The string is made of only two characters ‘a’ and ‘b’.
  3. You can change k characters.
    Example: string=“abab” and k=1.
    By changing the first b to a the string will become aaab
  4. You have to print the maximum length of a substring that contains same characters (either all a or all b) with k changes possible. Here the output will be 3.

You can 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

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

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Sanket and strings