How do I approach the Minimum Length of String After Deleting Similar Ends problem?

The problem can be efficiently solved using the two-pointer approach, where you compare the characters at both ends of the string and remove them if they are the same. This continues until no more operations are possible.

What is the time complexity of the solution?

The time complexity of the solution is O(n), where n is the length of the string. The two-pointer technique ensures that each character is checked once.

What are some common mistakes to avoid in this problem?

Avoid overcomplicating the solution with extra data structures or not managing the two pointers correctly. Also, ensure that edge cases are handled properly.

Can this problem be solved with a brute-force approach?

A brute-force approach could work but would be inefficient, as repeatedly removing characters from both ends could result in a time complexity worse than O(n). The two-pointer approach is optimal.

How does the two-pointer approach help in solving this problem?

The two-pointer approach helps efficiently check and remove characters from both ends of the string in linear time, reducing the problem to a manageable O(n) time complexity.

#1750

Medium

auto_awesome双·指针·invariant

LeetCode 题解工作台

删除字符串两端相同字符后的最短长度

给你一个只包含字符 'a' ， 'b' 和 'c' 的字符串 s ，你可以执行下面这个操作（5 个步骤）任意次：选择字符串 s 一个非空的前缀，这个前缀的所有字符都相同。选择字符串 s 一个非空的后缀，这个后缀的所有字符都相同。前缀和后缀在字符串中任意位置都不能有交集。前缀和后缀包含…

双指针字符串

题目描述

给你一个只包含字符 'a'，'b' 和 'c' 的字符串 s ，你可以执行下面这个操作（5 个步骤）任意次：

选择字符串 s 一个非空的前缀，这个前缀的所有字符都相同。
选择字符串 s 一个非空的后缀，这个后缀的所有字符都相同。
前缀和后缀在字符串中任意位置都不能有交集。
前缀和后缀包含的所有字符都要相同。
同时删除前缀和后缀。

请你返回对字符串 s 执行上面操作任意次以后（可能 0 次），能得到的 最短长度 。

示例 1：

输入：s = "ca"
输出：2
解释：你没法删除任何一个字符，所以字符串长度仍然保持不变。

示例 2：

输入：s = "cabaabac"
输出：0
解释：最优操作序列为：
- 选择前缀 "c" 和后缀 "c" 并删除它们，得到 s = "abaaba" 。
- 选择前缀 "a" 和后缀 "a" 并删除它们，得到 s = "baab" 。
- 选择前缀 "b" 和后缀 "b" 并删除它们，得到 s = "aa" 。
- 选择前缀 "a" 和后缀 "a" 并删除它们，得到 s = "" 。

示例 3：

输入：s = "aabccabba"
输出：3
解释：最优操作序列为：
- 选择前缀 "aa" 和后缀 "a" 并删除它们，得到 s = "bccabb" 。
- 选择前缀 "b" 和后缀 "bb" 并删除它们，得到 s = "cca" 。

提示：

1 <= s.length <= 10⁵
s 只包含字符 'a'，'b' 和 'c' 。

lightbulb

解题思路

方法一：双指针

我们定义两个指针 $i$ 和 $j$ 分别指向字符串 $s$ 的头部和尾部，然后向中间移动，直到 $i$ 和 $j$ 指向的字符不相等，此时 $\max(0, j - i + 1)$ 即为答案。

时间复杂度 $O(n)$ ，空间复杂度 $O(1)$ 。其中 $n$ 为字符串 $s$ 的长度。

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class Solution:
    def minimumLength(self, s: str) -> int:
        i, j = 0, len(s) - 1
        while i < j and s[i] == s[j]:
            while i + 1 < j and s[i] == s[i + 1]:
                i += 1
            while i < j - 1 and s[j - 1] == s[j]:
                j -= 1
            i, j = i + 1, j - 1
        return max(0, j - i + 1)

speed

复杂度分析

指标	值
时间	O(n)
空间	O(1)

psychology

面试官常问的追问

外企场景

question_mark
Check if the candidate efficiently uses the two-pointer approach to minimize the number of steps.
question_mark
Look for correct identification of the invariant and its application in halting further removals.
question_mark
Evaluate how well the candidate handles edge cases, such as strings that cannot be reduced further.

warning

常见陷阱

外企场景

error
Failing to correctly manage the two pointers, leading to an incorrect stopping condition.
error
Overcomplicating the solution by using extra data structures when a simple two-pointer approach suffices.
error
Not handling edge cases, such as strings where no removal is possible or when all characters are the same.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Handling larger inputs efficiently, ensuring the solution works for strings with up to 100,000 characters.
arrow_right_alt
Adapting the solution to allow for different operations at each end (e.g., only removing matching prefixes).
arrow_right_alt
Modifying the problem to allow removals from non-matching parts of the string rather than only from the ends.

help

常见问题

外企场景

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