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

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

Java Code
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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();
    }
}

C++ Code
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class Solution {
public:
    int percentageLetter(string s, char letter) {
        int cnt = 0;
        for (char& c : s) cnt += c == letter;
        return cnt * 100 / s.size();
    }
};

Go Code
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func percentageLetter(s string, letter byte) int {
	cnt := 0
	for i := range s {
		if s[i] == letter {
			cnt++
		}
	}
	return cnt * 100 / len(s)
}

TypeScript Code
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function percentageLetter(s: string, letter: string): number {
    let count = 0;
    let total = s.length;
    for (let i of s) {
        if (i === letter) count++;
    }
    return Math.floor((count / total) * 100);
}

Rust Code
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impl Solution {
    pub fn percentage_letter(s: String, letter: char) -> i32 {
        let mut count = 0;
        for c in s.chars() {
            if c == letter {
                count += 1;
            }
        }
        ((count * 100) / s.len()) as i32
    }
}