message: update rearrange-string-k-distance-apart.py
  a. use s intead of str (str is python keyword)
  b. using Counter is a bit more Pythonic
  c. Not only k == 0 but also k == 1 reuse given s
diff: @@ -34,22 +34,19 @@ class Solution(object):
  # Time: O(nlogc), c is the count of unique characters.
  # Space: O(c)
  -from collections import defaultdict
  +from collections import Counter
  from heapq import heappush, heappop
  class Solution2(object):
  - def rearrangeString(self, str, k):
  + def rearrangeString(self, s, k):
  """
  :type str: str
  :type k: int
  :rtype: str
  """
  - if k == 0:
  - return str
  -
  - cnts = defaultdict(int)
  - for c in str:
  - cnts[c] += 1
  + if k <= 1:
  + return s
  + cnts = Counter(s)
  heap = []
  for c, cnt in cnts.iteritems():
  heappush(heap, [-cnt, c])
  @@ -57,7 +54,7 @@ class Solution2(object):
  result = []
  while heap:
  used_cnt_chars = []
  - for _ in xrange(min(k, len(str) - len(result))):
  + for _ in xrange(min(k, len(s) - len(result))):
  if not heap:
  return ""
  cnt_char = heappop(heap)
  