Link: 132. Split Palindrome String II - LeetCode
Given a string s, please split it s into substrings so that each substring is a palindrome.
Returns the minimum number of divisions that meet the requirements .
Example 1:
Input: s = "aab" Output: 1 Explanation: It only takes one split to split s into two palindrome substrings ["aa", "b"].
Example 2:
Input: s = "a" Output: 0
Example 3:
Input: s = "ab" Output: 1
hint:
1 <= s.length <= 2000sConsists of only lowercase English letters
Idea:
Define dp[i] as the minimum number of splits required to split all s[0...i] into strings
Define isRound[i][j] to indicate whether the string s[j...i] is a string
State transition equation:
The first thing to make clear is that i walks from left to right and j walks from right to left;
When s[i] == s[j] and the substring (s[i+1,i-1] is a palindrome string, that is, isRound[i][j] = True or s[i..j] only one or two characters):
dp[i] = min(dp[i],dp[j-1]+1)
Why is dp[j-1]+1? This is to treat the string after j as a whole divisible palindrome string, and then cut it at j
One thing to note is that when it is successfully determined that the current string is a palindrome string, if j equals 0, it means that the entire character is a string, just set dp[i] = 0
class Solution:
def minCut(self, s: str) -> int:
n = len(s)
dp = [0 for i in range(n+1)]
isRound = [[False for i in range(n+1)] for j in range(n+1)]
for i in range(1,n):
dp[i] = i
for j in range(i,-1,-1):
if s[i] == s[j] and (isRound[j+1][i-1] or i-j<2):
isRound[j][i] = True
if j==0: dp[i] = 0
else : dp[i] = min(dp[i],dp[j-1]+1)
return dp[n-1]