LeetCode 2278. Percentage of Letter in String Solution in Java, C++, Python & More | Explanation + Code

Description

Given a string s and a character letter, return the percentage of characters in s that equal letter rounded down to the nearest whole percent.

 

Example 1:

Input: s = "foobar", letter = "o"
Output: 33
Explanation:
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
Explanation:
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.

Solutions

Solution 1: Counting

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

Time complexity is O(n), where n is the length of the string s. Space complexity is O(1).

PythonJavaC++GoTypeScriptRust

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

class Solution { public int percentageLetter(String s, char letter) { int cnt = 0; for (char c : s.toCharArray()) { if (c == letter) { ++cnt; } } return cnt * 100 / s.length(); } }(code-box)

class Solution { public: int percentageLetter(string s, char letter) { return 100 * ranges::count(s, letter) / s.size(); } };(code-box)

func percentageLetter(s string, letter byte) int { return strings.Count(s, string(letter)) * 100 / len(s) }(code-box)

function percentageLetter(s: string, letter: string): number { const count = s.split('').filter(c => c === letter).length; return Math.floor((100 * count) / s.length); }(code-box)

impl Solution { pub fn percentage_letter(s: String, letter: char) -> i32 { let count = s.chars().filter(|&c| c == letter).count(); (100 * count as i32 / s.len() as i32) as i32 } }(code-box)