2278. Percentage of Letter in String

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

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

Python3JavaC++GoTypeScriptRust

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

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();
    }
}

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

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

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

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
    }
}

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