Geting the start and end index of the first consecutive sequence of the same character in a python String

Question

I want to get the start and finish index of the first consecutive sequence of the same character string in python

'aaabca' -> (0, 2)

'helllooo' ->  (2, 4)

'hellooo' -> (2,3)

'abcd' -> (-1, -1)

is there a super clean way to accomplish that?

Answer 1

You could use a regex that find the repetition of a char ( (\\w)\\1+ ), then get the position of the match (using m.start() and m.end() )

values = ['aaabca', 'helllooo', 'hellooo', 'abcd']

for value in values:
    m = re.search(r'(\w)\1+', value)
    if m:
        print(f'{value:10s}{str((m.start(), m.end() - 1)):10s}{m.group(0)}')
    else:
        print(f'{value:10s}{str((-1, -1)):10s}')

Giving

aaabca    (0, 2)    aaa
helllooo  (2, 4)    lll
hellooo   (2, 3)    ll
abcd      (-1, -1)

---

Note

To change the type of char to search a repetition on, replace the \\w

• (\\d)\\1+ repetition of a digit
• (.)\\1+ repetition of any char
• ([az])\\1+ repetition of a lower case
• ...

Answer 2

Here's one way to do it

x = "helllooo"

count = 0
start = -1
end = -1
for i in range(len(x)-1):
    if x[i] == x[i+1]:
        if count == 0:
            start = i
        count += 1
        end = start + count
    else:
        if count > 0:
            break
        count = 0

print(start, end)

Answer 3

My first answer on stackoverflow. I hope you can give me rep :) Totally works have a nice day

word_list=['aaabca','helllooo','hellooo','abcd']

def find(word):
    first_char=[]
    index_list=[]

    for n,i in enumerate(word):
         
        if n+1<len(word):
            
            if i==word[n+1]:
                first_char.append(i)
                
                while first_char[0]==i:
                    
                    index_list.append(n)
                    index_list.append(n+1)
                    break
    try:
        print(index_list[0],index_list[-1])
    except:
        print(-1,-1)
    
for word in word_list:
    find(word)