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rearrange-string-k-distance-apart.py 1.77 KB
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Sanghee Kim 提交于 2019-02-03 22:34 +08:00 . update rearrange-string-k-distance-apart.py
# Time: O(n)
# Space: O(n)
class Solution(object):
def rearrangeString(self, str, k):
"""
:type str: str
:type k: int
:rtype: str
"""
cnts = [0] * 26
for c in str:
cnts[ord(c) - ord('a')] += 1
sorted_cnts = []
for i in xrange(26):
sorted_cnts.append((cnts[i], chr(i + ord('a'))))
sorted_cnts.sort(reverse=True)
max_cnt = sorted_cnts[0][0]
blocks = [[] for _ in xrange(max_cnt)]
i = 0
for cnt in sorted_cnts:
for _ in xrange(cnt[0]):
blocks[i].append(cnt[1])
i = (i + 1) % max(cnt[0], max_cnt - 1)
for i in xrange(max_cnt-1):
if len(blocks[i]) < k:
return ""
return "".join(map(lambda x : "".join(x), blocks))
# Time: O(nlogc), c is the count of unique characters.
# Space: O(c)
from collections import Counter
from heapq import heappush, heappop
class Solution2(object):
def rearrangeString(self, s, k):
"""
:type str: str
:type k: int
:rtype: str
"""
if k <= 1:
return s
cnts = Counter(s)
heap = []
for c, cnt in cnts.iteritems():
heappush(heap, [-cnt, c])
result = []
while heap:
used_cnt_chars = []
for _ in xrange(min(k, len(s) - len(result))):
if not heap:
return ""
cnt_char = heappop(heap)
result.append(cnt_char[1])
cnt_char[0] += 1
if cnt_char[0] < 0:
used_cnt_chars.append(cnt_char)
for cnt_char in used_cnt_chars:
heappush(heap, cnt_char)
return "".join(result)
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