Problem

Given a string s and a character letter, return _thepercentage of characters in _s that equalletter rounded down to the nearest whole percent.

Examples

Example 1

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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

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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

Method 1 - Count & Percentage

Intuition

Count the number of times letter appears in s, compute the percentage, and round down to the nearest integer.

Approach

Use a simple loop or built-in count to compute cnt = occurrences of letter in s, let n = len(s), then return cnt * 100 // n (integer division to get the floored percentage).

Code

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#include <string>
using namespace std;
class Solution {
public:
    int percentageLetter(string s, char letter) {
        int cnt = 0, n = s.size();
        for (char c : s) if (c == letter) cnt++;
        return cnt * 100 / n;
    }
};
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func percentageLetter(s string, letter byte) int {
    cnt := 0
    for i := 0; i < len(s); i++ {
        if s[i] == letter { cnt++ }
    }
    return cnt * 100 / len(s)
}
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class Solution {
    public int percentageLetter(String s, char letter) {
        int cnt = 0, n = s.length();
        for (char c : s.toCharArray()) if (c == letter) cnt++;
        return cnt * 100 / n;
    }
}
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class Solution {
    fun percentageLetter(s: String, letter: Char): Int {
        val cnt = s.count { it == letter }
        return cnt * 100 / s.length
    }
}
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def percentageLetter(s: str, letter: str) -> int:
    return s.count(letter) * 100 // len(s)
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fn percentage_letter(s: &str, letter: char) -> i32 {
    let cnt = s.chars().filter(|&c| c == letter).count();
    (cnt * 100 / s.len()) as i32
}
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function percentageLetter(s: string, letter: string): number {
    let cnt = 0;
    for (const c of s) if (c === letter) cnt++;
    return Math.floor(cnt * 100 / s.length);
}

Complexity

  • ⏰ Time complexity: O(n)
  • 🧺 Space complexity: O(1)