Leetcode 1278. Palindrome Partitioning III

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本文探讨了一道算法题目，旨在寻找将给定字符串分割为k个回文子串所需的最小字符更改数。通过动态规划算法，文章详细解释了如何高效地解决这一问题，并提供了具体的代码实现。

You are given a string s containing lowercase letters and an integer k. You need to :

First, change some characters of s to other lowercase English letters.
Then divide s into k non-empty disjoint substrings such that each substring is palindrome.

Return the minimal number of characters that you need to change to divide the string.

Example 1:

Input: s = "abc", k = 2
Output: 1
Explanation: You can split the string into "ab" and "c", and change 1 character in "ab" to make it palindrome.

Example 2:

Input: s = "aabbc", k = 3
Output: 0
Explanation: You can split the string into "aa", "bb" and "c", all of them are palindrome.

Example 3:

Input: s = "leetcode", k = 8
Output: 0

Constraints:

1 <= k <= s.length <= 100.
s only contains lowercase English letters.

Accepted

4,625

Submissions

8,039

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时间复杂度是O(N^2*k)，k是题目中的参数，很难优化下去了。各种边界条件是这个题目的难点：

class Solution:
    def palindromePartition(self, s, k):
        l = len(s)
        cost = [[0 for i in range(l)] for j in range(l)] #
        dp = [[l for j in range(k + 2)] for i in range(l)] #[0..i]的字符串分j次最少是多少，i是最后一个元素

        for span in range(2, l + 1):
            for i in range(0, l + 1 - span):
                start, end = i, i + span - 1
                same = 0 if s[start] == s[end] else 1
                cost[start][end] = cost[start + 1][end - 1] + same
        for i in range(l):
            dp[i][1] = cost[0][i]
            for j in range(2, min(i + 2, k + 2)): #bug1: min(i+1,k+1)
                for i0 in range(0, i):
                    dp[i][j] = min(dp[i0][j - 1] + cost[i0 + 1][i], dp[i][j])
        return dp[l - 1][k]

s = Solution()
print(s.palindromePartition("abc",2))
print(s.palindromePartition("aabbc",3))
print(s.palindromePartition("leetcode",8))