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palindrome-partitioning-iii.py 1.04 KB
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# Time: O(k * n^2)
# Space: O(n^2)
class Solution(object):
def palindromePartition(self, s, k):
"""
:type s: str
:type k: int
:rtype: int
"""
# dp1[i][j]: minimum number of changes to make s[i, j] palindrome
dp1 = [[0]*len(s) for _ in xrange(len(s))]
for l in xrange(1, len(s)+1):
for i in xrange(len(s)-l+1):
j = i+l-1
if i == j-1:
dp1[i][j] = 0 if s[i] == s[j] else 1
elif i != j:
dp1[i][j] = dp1[i+1][j-1] if s[i] == s[j] else dp1[i+1][j-1]+1
# dp2[d][i]: minimum number of changes to divide s[0, i] into d palindromes
dp2 = [[float("inf")]*len(s) for _ in xrange(2)]
dp2[1] = dp1[0][:]
for d in xrange(2, k+1):
dp2[d%2] = [float("inf")]*len(s)
for i in xrange(d-1, len(s)):
for j in xrange(d-2, i):
dp2[d%2][i] = min(dp2[d%2][i], dp2[(d-1)%2][j]+dp1[j+1][i])
return dp2[k%2][len(s)-1]
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