#2278

Easy

auto_awesomeString-driven solution strategy

LeetCode Problem Workspace

Percentage of Letter in String

Calculate the percentage of a specific letter in a string using a direct string-driven counting approach, rounding down correctly.

string

Problem Statement

You are given a string s and a lowercase character letter. Your task is to return the percentage of characters in s that are equal to letter, rounded down to the nearest whole percent.

For example, if s = "foobar" and letter = "o", the output should be 33 because 2 of 6 characters match 'o'. Handle cases where the letter does not appear at all and ensure all calculations follow integer rounding rules.

Examples

Example 1

Input: s = "foobar", letter = "o"

Output: 33

The percentage of characters in s that equal the letter 'o' is 2 / 6 * 100% = 33% when rounded down, so we return 33.

Example 2

Input: s = "jjjj", letter = "k"

Output: 0

The percentage of characters in s that equal the letter 'k' is 0%, so we return 0.

Constraints

1 <= s.length <= 100
s consists of lowercase English letters.
letter is a lowercase English letter.

Solution Approach

Count Occurrences of Letter

Iterate through the string s and count each occurrence of the target letter. This is the core string-driven strategy to directly quantify matches without extra data structures.

Calculate Percentage

Divide the letter count by the length of s, multiply by 100, and use floor division or integer cast to round down. This ensures the returned value adheres to the problem's rounding requirement.

Return Result

Return the computed integer percentage. Edge cases include when the letter is absent (return 0) or when the entire string matches the letter (return 100).

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time 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.

What Interviewers Usually Probe

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

Common Pitfalls or Variants

Common pitfalls

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

Follow-up variants

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

FAQ

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.

terminal

Solution

Solution 1: Counting

We can traverse the string $\textit{s}$ and count the number of characters that are equal to $\textit{letter}$. Then, we calculate the percentage using the formula $\textit{count} \times 100 \, / \, \textit{len}(\textit{s})$.

1

2

3

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

Continue Practicing

#2273 find resultant array after removing anagrams #2283 check if number has equal digit count and digit value #2284 sender with largest word count