CODEWITHSUNDEEP
stringpython-3.6sequence
Jay
Jay

Reputation: 41

“Is there a way to get sequence of most consecutive alphabet?”

I am working on a python problem which has a string such as “aaabbcc” and a number n (integer). I have to display a sequence of any Alphabetic character that appears exactly n times.

I have tried the code

import collections
str1 = 'aaabbcc'
d = collections.defaultdict(int)
for c in str1:
    d[c] += 1

for c in sorted(d, key=d.get, reverse=True):
  if d[c] > 1:
      print(c, d[c])

But I am getting output as 

a 3
b 2
c 2

I am expecting output as the integer input 3 is input is taken, from user. 

integer= 3 
sequence= aaa

Is there any alternative solution?

Upvotes: 2

Views: 52

Answers (3)

Tim Biegeleisen
Tim Biegeleisen

Reputation: 521279

Here is a regex based approach which seems to work:

input = "ddaaabbbbbbbbccceeeeeee"
n = 3
for match in re.finditer(r'(.)(?!\1)(.)\2{' + str(n-1) + r'}(?!\2)', input):
print(match.group(0)[1:])

aaa
ccc

The regex pattern being used in the exact example above is this:

(.)(?!\1)(.)\2{2}(?!\2)

This says to:

(.)     match and capture any single character
(?!\1)  assert that the next character is different
(.)     then match and capture that next character
\2{2}   which is then followed by that same character exactly twice (total of 3)
(?!\2)  after three instances, the character that follows is NOT the same

Upvotes: 1

hiro protagonist
hiro protagonist

Reputation: 46859

a loop-based approach (that should be pretty straight-forward):

str1 = 'aaabbcc'
n = 3

count = 1
last = None
for char in str1:
    if last == char:
        count += 1
    else:
        if count == n:
            print(f'integer: {n} sequence: {n*last}')
        last = char
        count = 1
if count == n:
    print(f'integer: {n} sequence: {n*last}')

the last if statement is there to print a solution if one was found including the last character of str1.

Upvotes: 0

hiro protagonist
hiro protagonist

Reputation: 46859

an itertools.groupby-based approach:

from itertools import groupby

str1 = 'aaabbcc'
n = 3

for key, group in groupby(str1):
    if len(tuple(group)) == n:
        print(f'integer: {n} sequence: {n*key}')

withot a key groupby will group the sequence by identity - i.e. every time the letter in str1 changes it will yield that letter and its occurrences.

Upvotes: 1

Related Questions