Welcome to Subscribe On Youtube

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

  • 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) {
            int cnt = 0;
            for (char& c : s) cnt += c == letter;
            return cnt * 100 / s.size();
        }
    };
    
  • class Solution:
        def percentageLetter(self, s: str, letter: str) -> int:
            return s.count(letter) * 100 // len(s)
    
    
  • 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
        }
    }
    
    

All Problems

All Solutions