Rearrange String k Distance Apart
Question
Given a string s and an integer k, rearrange the string such that every substring of length k is lexicographically ordered and the concatenation of all the substrings is equal to s.
Example 1
None
Solution
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all//Rearrange String k Distance Apart.py
import collections
def rearrangeStringKDistanceApart(s, k):
if k == 0:
return s
max_freq = collections.Counter(s).most_common(1)[0][1]
max_char = collections.Counter(s).most_common(1)[0][0]
if (len(s) < (max_freq - 1) * (k + 1) + 1):
return ""
res = [max_char] * max_freq
used = [False] * len(s)
for i in range(max_freq):
pointer = i
while used[pointer]:
pointer += 1
res[i] = s[pointer]
used[pointer] = True
pointer = max_freq
for i in range(len(s)):
if s[i] != max_char and not used[i]:
while used[pointer]:
pointer += 1
if pointer - i > k:
return ""
res.insert(pointer, s[i])
used[pointer] = True
pointer += 1
return "".join(res)
# Test
s = "aabbcc"
k = 3
print(rearrangeStringKDistanceApart(s, k))
# Output: ababcc
all//Rearrange String k Distance Apart.py
import collections
def rearrangeStringKDistanceApart(s, k):
if k == 0:
return s
max_freq = collections.Counter(s).most_common(1)[0][1]
max_char = collections.Counter(s).most_common(1)[0][0]
if (len(s) < (max_freq - 1) * (k + 1) + 1):
return ""
res = [max_char] * max_freq
used = [False] * len(s)
for i in range(max_freq):
pointer = i
while used[pointer]:
pointer += 1
res[i] = s[pointer]
used[pointer] = True
pointer = max_freq
for i in range(len(s)):
if s[i] != max_char and not used[i]:
while used[pointer]:
pointer += 1
if pointer - i > k:
return ""
res.insert(pointer, s[i])
used[pointer] = True
pointer += 1
return "".join(res)
# Test
s = "aabbcc"
k = 3
print(rearrangeStringKDistanceApart(s, k))
# Output: ababcc