Description
Given a string s, partition s such that every substring of the partition is a palindrome.
Return the minimum cuts needed for a palindrome partitioning of s.
Example:
Input: "aab"
Output: 1
Explanation: The palindrome partitioning ["aa","b"] could be produced using 1 cut.
Solutions
1. DP
a b a | c c
j i
j-1 | [j, i] is palindrome
dp(j-1) + 1
- 如果字串 [j, i] 是回文串，那么有 dp[i] = min(dp[j - 1] + 1)，其中 (j <= i)
- 还有一个值得注意的是，如果字串 [j, i] 是回文串，那么 [j + 1, i - 1] 也是回文串, 并且 c[j] == c[i]。
# Time: O(n^2)
# Space: O(n)
class Solution:
def minCut(self, s: str) -> int:
if not s:
return 0
n = len(s)
dp = [0 for _ in range(n)]
is_pal = [[False for _ in range(n)] for _ in range(n)]
for i in range(n):
min_cuts = i
j = 0
while j <= i:
if s[j] == s[i] and (i - j < 2 or is_pal[j + 1][i - 1]):
is_pal[j][i] = True
min_cuts = 0 if j == 0 else min(min_cuts, dp[j - 1] + 1)
j += 1
dp[i] = min_cuts
return dp[-1]
# 29/29 cases passed (540 ms)
# Your runtime beats 41.7 % of python3 submissions
# Your memory usage beats 80 % of python3 submissions (31.1 MB)