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