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  1. Prepare
  2. Python
  3. Itertools
  4. Compress the String!
  5. Discussions

Compress the String!

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860 Discussions

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  • 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))

  • vyoma_patel
     + 0 comments
    from itertools import groupby
s = input()
for char, grouped_iter in groupby(s):
    print(f"({len(list(grouped_iter))}, {char})", end = " ")

  • msh_123
     + 1 comment

    from itertools import groupby

    input_value = input()

    group_result = groupby(input_value)

    for key,value in group_result:

    value = sorted(list(value))
print(f"({int(len(value))}, {int(key)})", end=" ")

  • rohitambasta_ii1
     + 0 comments

    Core Python code (Without Built-in function)- We will appreciate itertools module some other time :

    string= input()

lst= [element for element in string]

count= 1

prev= lst[0]

for i in range(1, len(lst)):
    if lst[i] == prev:
        count = count + 1

    else:
        print(f"({count}, {prev})", end= " ")
        prev= lst[i]
        count = 1

print(f"({count}, {prev})")