Download Palindrome Partitioning II.py
class Solution(object): # def minCut(self, s): # """ # :type s: str # :rtype: int # """ # # https://discuss.leetcode.com/topic/2840/my-solution-does-not-need-a-table-for-palindrome-is-it-right-it-uses-only-o-n-space ls = len(s) cut = [i -1 for i in range(ls + 1)] for i in range(ls): # odd length pos = 0 while i - pos >= 0 and i + pos < ls and s[i - pos] == s[i + pos]: cut[i + pos + 1] = min(cut[i + pos + 1], 1 + cut[i - pos]) pos += 1 # even length pos = 1 while i - pos + 1 >= 0 and i + pos < ls and s[i - pos + 1] == s[i + pos]: cut[i + pos + 1] = min(cut[i + pos + 1], 1 + cut[i - pos + 1]) pos += 1 return cut[ls] # def minCut(self, s): # if len(s) <= 1: # return 0 # ls = len(s) # d =  * ls # pal = [[False] * ls for _ in range(ls)] # for i in range(ls - 1, -1, -1): # d[i] = ls - i - 1 # for j in range(i, ls): # if s[i] == s[j] and (j - i < 2 or pal[i + 1][j - 1]): # pal[i][j] = True # if j == ls - 1: # d[i] = 0 # elif d[j + 1] + 1 < d[i]: # d[i] = d[j + 1] + 1 # return d
Leetcode 132 Palindrome Partitioning II problem solution in python3 with explanation. This is the best place to expand your knowledge and get prepared for your next interview.
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