How do hash tables help in solving the Minimum Length of String After Operations problem?

Hash tables store the frequency of each character, which is essential for determining if adjacent pairs can be removed. This allows for an efficient solution.

What are the time and space complexities for this problem?

The time complexity is O(n), and the space complexity is O(1), as only a fixed amount of space is needed for frequency counting.

What’s the key to efficiently solving this problem?

The key is focusing on character frequencies and using a hash table to track them, minimizing the string through pair removal based on these frequencies.

Can this problem be solved using a stack approach?

Yes, a stack could potentially be used to remove adjacent pairs, but a hash table-based approach is more efficient in terms of frequency counting.

How do I handle edge cases in the Minimum Length of String After Operations?

Edge cases include strings that already have no adjacent pairs or strings where characters cannot fully form pairs. Ensure you account for these scenarios during implementation.

#3223

Medium

auto_awesome哈希·表·结合·string

LeetCode 题解工作台

操作后字符串的最短长度

给你一个字符串 s 。你需要对 s 执行以下操作任意次：选择一个下标 i ，满足 s[i] 左边和右边都至少有一个字符与它相同。删除 i 左边离它最近的 s[i] 字符。删除 i 右边离它最近的 s[i] 字符。请你返回执行完所有操作后， s 的最短长度。示例 1…

哈希表字符串计数

题目描述

给你一个字符串 s 。

你需要对 s 执行以下操作任意次：

选择一个下标 i ，满足 s[i] 左边和右边都至少有一个字符与它相同。
删除 i 左边离它最近的 s[i] 字符。
删除 i 右边离它最近的 s[i] 字符。

请你返回执行完所有操作后， s 的最短长度。

示例 1：

输入：s = "abaacbcbb"

输出：5

解释：
我们执行以下操作：

选择下标 2 ，然后删除下标 0 和 3 处的字符，得到 s = "bacbcbb" 。
选择下标 3 ，然后删除下标 0 和 5 处的字符，得到 s = "acbcb" 。

示例 2：

输入：s = "aa"

输出：2

解释：
无法对字符串进行任何操作，所以返回初始字符串的长度。

提示：

1 <= s.length <= 2 * 10⁵
s 只包含小写英文字母。

lightbulb

解题思路

方法一：计数

我们可以统计字符串中每个字符出现的次数，然后遍历计数数组，如果字符出现的次数为奇数，则最后该字符剩余 $1$ 个，如果字符出现的次数为偶数，则最后该字符剩余 $2$ 个。我们可以将所有字符的剩余个数相加，即为最终字符串的长度。

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class Solution:
    def minimumLength(self, s: str) -> int:
        cnt = Counter(s)
        return sum(1 if x & 1 else 2 for x in cnt.values())

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
Can the candidate quickly implement a solution using a hash table to count frequencies?
question_mark
Does the candidate consider all edge cases like an already minimal string or no removable pairs?
question_mark
Is the candidate aware of the importance of focusing only on character frequencies for an optimized solution?

warning

常见陷阱

外企场景

error
Ignoring edge cases where no adjacent pairs exist and not optimizing for a direct frequency-based solution.
error
Incorrectly handling cases with an odd number of identical characters that can’t fully form pairs.
error
Overcomplicating the solution by attempting to manually remove pairs instead of focusing on character frequencies.

swap_horiz

进阶变体

外企场景

arrow_right_alt
What happens if the string is composed entirely of unique characters?
arrow_right_alt
Can this problem be solved using a stack approach instead of hash tables?
arrow_right_alt
How does the time complexity change when considering strings of varying lengths or different characters?

help

常见问题

外企场景

继续练习

#3144 分割字符频率相等的最少子字符串 #3138 同位字符串连接的最小长度 #3137 K 周期字符串需要的最少操作次数