Hard
Make String Anti-palindrome — Python
Full explanation · Time O(n + 26) · Space O(26)
# Time: O(n + 26)
# Space: O(26)
# counting sort, greedy
class Solution(object):
def makeAntiPalindrome(self, s):
"""
:type s: str
:rtype: str
"""
cnt = [0]*26
for x in s:
cnt[ord(x)-ord('a')] += 1
if max(cnt) > len(s)//2:
return "-1"
result = [i for i, x in enumerate(cnt) for _ in xrange(x)]
l = next(l for l in xrange((len(s)//2)//2+1) if result[len(s)//2+l] != result[len(s)//2-1])
if l:
for i in xrange(cnt[result[len(s)//2-1]]-l):
result[len(s)//2+i], result[len(s)//2+i+l] = result[len(s)//2+i+l], result[len(s)//2+i]
return "".join(map(lambda x: chr(ord('a')+x), result))
# Time: O(n + 26)
# Space: O(26)
# counting sort, greedy, two pointers
class Solution2(object):
def makeAntiPalindrome(self, s):
"""
:type s: str
:rtype: str
"""
cnt = [0]*26
for x in s:
cnt[ord(x)-ord('a')] += 1
if max(cnt) > len(s)//2:
return "-1"
result = [i for i, x in enumerate(cnt) for _ in xrange(x)]
left = len(s)//2
right = left+1
while right < len(s) and result[right] == result[left]:
right += 1
while result[left] == result[len(s)-1-left]:
result[left] , result[right] = result[right], result[left]
left += 1
right += 1
return "".join(map(lambda x: chr(ord('a')+x), result))
# Time: O(n * 26)
# Space: O(26)
# freq table, greedy
class Solution3(object):
def makeAntiPalindrome(self, s):
"""
:type s: str
:rtype: str
"""
cnt = [0]*26
for x in s:
cnt[ord(x)-ord('a')] += 1
if max(cnt) > len(s)//2:
return "-1"
result = [-1]*len(s)
for i in xrange(len(s)//2):
j = next(j for j in xrange(len(cnt)) if cnt[j])
cnt[j] -= 1
result[i] = j
for i in xrange(len(s)//2, len(s)):
j = next(j for j in xrange(len(cnt)) if cnt[j] and result[(len(s)-1)-i] != j)
cnt[j] -= 1
result[i] = j
return "".join(map(lambda x: chr(ord('a')+x), result))