Leetcode 3563. Lexicographically Smallest String After Adjacent Removals

Espresso Macchiato2025-05-28 1:03

[Leetcode 3563. Lexicographically Smallest String After Adjacent Removals](#Leetcode 3563. Lexicographically Smallest String After Adjacent Removals)
- [1. 解题思路](#1. 解题思路)
- [2. 代码实现](#2. 代码实现)

题目链接：3563. Lexicographically Smallest String After Adjacent Removals

1. 解题思路

这次的最后一题同样没有自力搞定，简直了......

这道题还是一个动态规划的题目，就是提前构建一下消解空间，给定一个 N 2 N^2 N2的矩阵dp，其中任意元素dp[i][j]表示子串s[i:j+1]是否可以被可以被完全remove掉。

然后，我们就只需要考察一下从尾部到头部每一个元素被消除以及不被消除的两种情况下所能获得的最优解即可。

2. 代码实现

给出python代码实现如下：

python 复制代码

class Solution:
    def lexicographicallySmallestString(self, s: str) -> str:
        n = len(s)
        remEmpty = [[False] * n for _ in range(n)]
        
        def is_consecutive(a, b):
            d = abs(ord(a) - ord(b))
            return d == 1 or d == 25
        
        for i in range(n - 1):
            if is_consecutive(s[i], s[i + 1]):
                remEmpty[i][i + 1] = True
        for L in range(4, n + 1, 2):
            for i in range(n - L + 1):
                j = i + L - 1
                for k in range(i + 1, j + 1, 2):
                    if is_consecutive(s[i], s[k]) and (k == i + 1 or remEmpty[i + 1][k - 1]) and (k == j or remEmpty[k + 1][j]):
                        remEmpty[i][j] = True
                        break
        f = [""] * (n + 1)
        for i in range(n - 1, -1, -1):
            best = s[i] + f[i + 1]
            for j in range(i + 1, n, 2):
                if remEmpty[i][j] and f[j + 1] < best:
                    best = f[j + 1]
            f[i] = best
        return f[0]

提交代码评测得到：耗时6771ms，占用内存18.5MB。