Question

How to replace the pattern in the string with

     decoded_str=" Name(++info++)Age(++info++)Adress of the emp(++info++)"

 The first pattern "(++info++)" needs to replaced with (++info a++)
 The second pattern "(++info++)" needs to replaced with (++info b++)
 The third pattern "(++info++)" needs to replaced with (++info c++)
 If there many more then it should be replaced accordingly

Answer 1

This should be simple enough:

for character in range(ord('a'), ord('z')):
    if "(++info++)" not in decoded_str:
        break
    decoded_str = decoded_str.replace("(++info++)", "(++info {0}++)".format(chr(character)), 1)

print decoded_str

It has the added benefit of stopping at ‘z’. If you want to wrap around:

import itertools

for character in itertools.cycle(range(ord('a'), ord('z'))):
    if "(++info++)" not in decoded_str:
        break
    decoded_str = decoded_str.replace("(++info++)", "(++info {0}++)".format(chr(character)), 1)

print decoded_str

And just for fun, a one-liner, and O(n):

dstr = "".join(x + "(++info {0}++)".format(chr(y)) for x, y in zip(dstr.split("(++info++)"), range(ord('a'), ord('z'))))[:-len("(++info a++)")]