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

|

863 Discussions

|

  • HarshPrashar1
     + 0 comments

    from itertools import groupby

    def run_length_encoding(s): encoded_parts = [] for key, group in groupby(s): count = len(list(group)) encoded_parts.append(f'({count}, {key})') return ' '.join(encoded_parts)

    input_string = input() output_string = run_length_encoding(input_string) print(output_string)

  • pranaynikhare7
     + 0 comments
    s = input()

for  key, grp in groupby(s):
    print(f"({len(list(grp))}, {key})", end=" ")

  • maluisamrok
     + 0 comments
    import itertools
print (*((len(list(g)), int(k)) for k,g in itertools.groupby(input())))

  • lsloan
     + 0 comments

    My shortest solution…

    from itertools import *
print(*list((len(list(g)), int(k)) for k, g in groupby(input())))

    A solution without groupby()…

    o = p = []
for c in map(int, input()):
  if c == p:
    n += 1
    continue
  o += [(n, p)] if p != [] else []
  p = c
  n = 1
print(*(o + [(n, p)]))

  • dilajevolves
     + 0 comments
    from itertools import groupby
result = [f"({len(list(group))}, {char})" for char,group in groupby(input())]
print (" ".join(result))