What is the fastest way to count letter occurrences in a string?

Iterate through the string once, incrementing a counter when a character matches the target letter. This O(n) approach is simple and direct.

How do I ensure the percentage is rounded down?

Use integer division or a floor function after multiplying by 100 to round down automatically.

What if the letter does not appear in the string?

Return 0 because no characters match the target letter, following the problem requirement for rounding.

Can this solution handle strings of length 1?

Yes, the same counting and percentage calculation works, returning either 0 or 100 as appropriate.

Is this problem pattern considered a string-driven solution strategy?

Yes, it relies entirely on scanning the string and counting target characters, exemplifying the string-driven pattern.

#2278

Easy

auto_awesomeString-driven solution strategy

LeetCode 题解工作台

字母在字符串中的百分比

给你一个字符串 s 和一个字符 letter ，返回在 s 中等于 letter 字符所占的百分比，向下取整到最接近的百分比。示例 1：输入： s = "foobar", letter = "o" 输出： 33 解释：等于字母 'o' 的字符在 s 中占到的百分比是 2 / 6 * 100…

字符串

题目描述

给你一个字符串 s 和一个字符 letter ，返回在 s 中等于 letter 字符所占的 百分比 ，向下取整到最接近的百分比。

示例 1：

输入：s = "foobar", letter = "o"
输出：33
解释：
等于字母 'o' 的字符在 s 中占到的百分比是 2 / 6 * 100% = 33% ，向下取整，所以返回 33 。

示例 2：

输入：s = "jjjj", letter = "k"
输出：0
解释：
等于字母 'k' 的字符在 s 中占到的百分比是 0% ，所以返回 0 。

提示：

1 <= s.length <= 100
s 由小写英文字母组成
letter 是一个小写英文字母

lightbulb

解题思路

方法一：计数

我们可以遍历字符串 $\textit{s}$ ，统计其中等于 $\textit{letter}$ 的字符的个数，然后根据公式 $\textit{count} \times 100 \, / \, \textit{len}(\textit{s})$ 计算百分比。

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

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class Solution:
    def percentageLetter(self, s: str, letter: str) -> int:
        return s.count(letter) * 100 // len(s)

speed

复杂度分析

指标	值
时间	complexity is O(n) where n is the length of s because each character is inspected once. Space complexity is O(1) since only counters are used, independent of input size.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Can you count how many times the letter appears in s efficiently?
question_mark
Do you handle rounding down correctly for percentages less than 1?
question_mark
How do you handle the case when the letter does not appear at all?

warning

常见陷阱

外企场景

error
Forgetting to round down and returning a float or rounded value incorrectly.
error
Miscounting letter occurrences in strings with repeated characters.
error
Not handling empty or single-character strings according to constraints.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Return the percentage as a floating-point number instead of rounding down.
arrow_right_alt
Compute the percentage of multiple letters at once in the same string.
arrow_right_alt
Handle uppercase letters and case-insensitive matching.

help

常见问题

外企场景

继续练习

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