What is the main strategy for solving 'Count Asterisks'?

The strategy is to iterate through the string while tracking whether you're inside or outside a pair of '|'. Only count '*' characters outside those pairs.

How do you track whether you're inside a pair of '|' characters?

You can use a boolean flag that toggles each time a '|' character is encountered. This helps determine whether a '*' is inside or outside the pair.

Can I use a stack to solve the 'Count Asterisks' problem?

While you could use a stack, a simpler and more efficient approach is to use a boolean flag to track whether you're inside a '|'-pair.

What is the time complexity of this problem?

The time complexity is O(n), where n is the length of the string, as the algorithm only needs to iterate through the string once.

Are there any edge cases I should consider for this problem?

Yes, make sure to handle cases where there are no '*' characters, or where '*' characters appear in unusual patterns, such as consecutive '*' outside of '|' pairs.

#2315

Easy

auto_awesomeString-driven solution strategy

LeetCode 题解工作台

统计星号

字符串

题目描述

给你一个字符串 s ，每两个连续竖线 '|' 为一对。换言之，第一个和第二个 '|' 为一对，第三个和第四个 '|' 为一对，以此类推。

请你返回不在竖线对之间，s 中 '*' 的数目。

注意，每个竖线 '|' 都会恰好属于一个对。

示例 1：

输入：s = "l|*e*et|c**o|*de|"
输出：2
解释：不在竖线对之间的字符加粗加斜体后，得到字符串："l|*e*et|c**o|*de|" 。
第一和第二条竖线 '|' 之间的字符不计入答案。
同时，第三条和第四条竖线 '|' 之间的字符也不计入答案。
不在竖线对之间总共有 2 个星号，所以我们返回 2 。

示例 2：

输入：s = "iamprogrammer"
输出：0
解释：在这个例子中，s 中没有星号。所以返回 0 。

示例 3：

输入：s = "yo|uar|e**|b|e***au|tifu|l"
输出：5
解释：需要考虑的字符加粗加斜体后："yo|uar|e**|b|e***au|tifu|l" 。不在竖线对之间总共有 5 个星号。所以我们返回 5 。

提示：

1 <= s.length <= 1000
s 只包含小写英文字母，竖线 '|' 和星号 '*' 。
s 包含偶数个竖线 '|' 。

lightbulb

解题思路

方法一：模拟

我们定义一个整型变量 $\textit{ok}$ ，表示遇到 * 时是否能计数，初始时 $\textit{ok}=1$ ，表示可以计数。

遍历字符串 $s$ ，如果遇到 *，则根据 $\textit{ok}$ 的值决定是否计数，如果遇到 |，则 $\textit{ok}$ 的值取反。

最后返回计数的结果。

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

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class Solution:
    def countAsterisks(self, s: str) -> int:
        ans, ok = 0, 1
        for c in s:
            if c == "*":
                ans += ok
            elif c == "|":
                ok ^= 1
        return ans

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Checks whether the candidate can efficiently handle string traversal.
question_mark
Assesses whether the candidate correctly identifies how to track the status of being inside or outside the '|' pair.
question_mark
Evaluates the candidate's understanding of edge cases and proper exclusion of characters between '|' pairs.

warning

常见陷阱

外企场景

error
Failing to handle the toggling of the inside/outside state when encountering '|' characters.
error
Counting '*' characters within a '|' pair, even though they should be excluded.
error
Not accounting for edge cases where there are no '*' characters or when they appear in odd patterns.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Variant with a larger string where performance optimization is required.
arrow_right_alt
Variant where the string includes other symbols along with '*' and '|', introducing more complexity in exclusion logic.
arrow_right_alt
Variant where the '|' characters are nested, requiring more careful tracking of their pairs.

help

常见问题

外企场景

继续练习

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