Minimum Characters required to make a String Palindromic

'''
We can solve this problem efficiently in O(n) time and O(1) space.
Approach : Take left pointer as 0 and right and n-1 and compare both characters
If both are same left++ and right–;
else
we can think that we have to add N-right elements at front to make last (N-right) characters palindrome so added = (n-right)
Now left again point to 0 as we have to check the string from 0th index .
but while the character at right position is same as 0th index char then we can add one less character at front  and left will increase by 1.
'''

class Solution:
    # @param A : string
    # @return an integer
    def solve(self, a):
        n=len(a)
        l,r=0,n-1
        added =0
        while l<r:
            if a[l]==a[r]:
                l,r=l+1,r-1
            else:
                l=0
                added=n-r #we will add a[r:][::-1] to start
                while a[l]==a[r]:#this while loop decreases the redundant characters that we added which already useful in making palindrome
                    l=l+1
                    added-=1
                r-=1#Finally we should reduce the actual r because we have added sufficent chracters from back

        return added
 '''
 e.g. a c a a
      |     |
      a c a a
        | |
      a a =a c a a  #Before equal we added required chraracters to make it palindrome
           |   |

      a a =a   c   a a
        x     | |
      final a a c a a

      added =1
 '''