Today's Freebie. It's marked as easy.
This problem was asked by Amazon.
Run-length encoding is a fast and simple method of encoding strings. The basic idea is to represent repeated successive characters as a single count and character. For example, the string "AAAABBBCCDAA" would be encoded as "4A3B2C1D2A".
Implement run-length encoding and decoding. You can assume the string to be encoded have no digits and consists solely of alphabetic characters. You can assume the string to be decoded is valid.
Another Python Using Generators
"""Example using generators for run length encode/decode."""
def rleunits(str1):
"""Generate run length encoded units."""
current = None
count = 0
for letter in str1:
if letter == current:
count += 1
else:
if current:
yield str(count)+current
count = 1
current = letter
def rldunits(str1):
"""Generate run length decoded units."""
count = None
for c in str1:
if c.isdigit():
if not count:
count = c
else:
count += c
else:
yield(c * int(count))
count = None
def main():
"""Drive the example."""
str1 = "AAAABBBBCDDEEFFFFFFFFFFF"
encoded = ''.join(rleunits(str1))
print(encoded)
unencoded = ''.join(rldunits(encoded))
print(unencoded)
main()
@Absinthe I hire you to implement it for me. :D
@Surasanji as soon as you send the check I will have a version with your name on it.. This one looks really easy though, the trick is making it pythonic. I can almost imagine a regex solution.
Basic non regex solution
@Absinthe@qoto.org
def encode(input):
out = ""
let = input[0]
num = 1
for i in range(1, len(input)):
if input[i] == input[i-1]:
num = num + 1
else:
out = out + str(num) + input[i-1]
let = input[i]
num = 1
out = out + str(num) + input[len(input)-1]
return out
def decode(input):
out = ""
i = 0
while i != (len(input)):
for p in range(0, int(input[i])):
out = out + input[i+1]
i = i + 2
return out
Basic non regex solution
@matrix07012 not sure how you feel about criticism, so I will hold back too much opinion. However, line 4 and 12 are unused.
Basic non regex solution
@Absinthe@qoto.org No problem, I looked up the proper solution now and mine is clumsier than I thought.
Basic non regex solution
@matrix07012 I just watched a great video on looping and iterators and perhaps I am a bit more critical of things:
https://www.youtube.com/watch?v=EnSu9hHGq5o
It might change your life, it will change your code!
BTW, did you check out my generator version?
Basic non regex solution
@Absinthe@qoto.org I just did, pretty interesting.
SPOILER ONE-LINER SOLUTIONS IN PYTHON
import re
encooded = "14A3B2C1D2A"
decoded = ""
# first attempt at decoding
list1 = re.findall('\d+\D', encooded)
for l in list1:
count, char = int(l[0:-1]) , l[-1]
decoded += ''.join(count * char)
print(decoded)
# one-liner using regex for decoding!
print(''.join([ int(l[0:-1]) * l[-1] for l in re.findall('\d+\D', encooded) ]))
# one liner using regex for encoding
print( ''.join([ str(len(match[1])+1) + match[0] for match in re.findall(r"(.)(\1*)", decoded)]))