We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Prepare
  2. Python
  3. Itertools
  4. Compress the String!
  5. Discussions

Compress the String!

Sort by

recency

|

816 Discussions

|

  • JanekPython09
     + 0 comments
    from itertools import groupby 
for N, group in groupby(input()): 
    print((len(list(group)), int(N)), end=' ')

  • rubinaParveen
     + 0 comments
    from itertools import groupby
s = input()
group = groupby(s,int)

for key, grp in group:
    print((len(list(grp)),key),end=" ")

  • aditaima2004
     + 0 comments

    My solution:-

    from itertools import groupby
S=input()
print(*[(len(tuple(elements)), key) for key, elements in groupby(S, int)])

  • soullamiayyoub
     + 0 comments

    Sometimes you need to try to solve the problem with your onw logic. and here is mine, , whats do you think champs ? 

    s = '1222311'
count = 1
for i in range(1, len(s)):
    # if i+1 < len(s):
    # print(i, i-1)
    if s[i] == s[i-1]:
        count += 1
    else: 
        print(f"({count}, {s[i-1]})", end=' ')
        count = 1
print(f"({count}, {s[-1]})")

  • biumahmud2012
     + 0 comments

    from itertools import groupby

    Reading input

    s = input()

    Using groupby to group consecutive identical characters

    result = [(len(list(group)), int(key)) for key, group in groupby(s)]

    Printing the result with the required format

    print(*result)