Formatted question description: https://leetcode.ca/all/2278.html

2278. Percentage of Letter in String

• Difficulty: Easy.
• Related Topics: String.
• Similar Questions: Sort Characters By Frequency.

Problem

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.

Solution (Java, C++, Python)

• class Solution {
public int percentageLetter(String s, char letter) {
int count = 0;
int n = s.length();
for (int i = 0; i < n; i++) {
if (s.charAt(i) == letter) {
++count;
}
}
return (count * 100) / (n);
}
}

############

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

############

# 2278. Percentage of Letter in String
# https://leetcode.com/problems/percentage-of-letter-in-string/

class Solution:
def percentageLetter(self, s: str, letter: str) -> int:
n = len(s)
c = s.count(letter)

return int(c / n * 100)


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

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

• 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);
}


• 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
}
}



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).