A solution to a codefights daily puzzle that was amusing and might be useful to reference

expandIt.py

# You are given a string s composed of letters and numbers, which was compressed with some algorithm. Every letter in s is followed by a number (possibly with leading zeros), which represents the number of times this letter occurs consecutively. For example, "aaaaaaaabbbbbbcc" would be given as "a8b6c2".

# Decompress s, sort its letters and return the kth (1-based) letter of the obtained string.

# Note: each letter occurs in s no more than 251 times.

# Example

# Expand_It("a2b3c2a1", 2) = "a"

# The decompressed s equals "aabbbcca", and once sorted it becomes "aaabbbcc". Its second 1-based letter is 'a', which is the answer.


import itertools

def Expand_It(s, k):
	if k <= 0:
		return -1
	fixed_list = [(key, sum(int(r[1]) for r in rows)) 
		for	key, rows 
		in itertools.groupby(sorted([(x,y) 
		for x, y 
			in pairwise(re.findall('[a-zA-Z]+|\\d+', s.strip(" ")))], key=lambda t: t[0]), key=lambda t: t[0])]
	if k <= sum(k for j,k in fixed_list):
		for pair in fixed_list:
			if k <= pair[1]:
				return pair[0]
			else:
				k -= pair[1]
	else:
		return -1