carloscn
commented
1 year ago Analysis
static int32_t percentage_letter(const char *s, char letter)
{
int32_t ret = -1;
size_t len;
UTILS_CHECK_PTR(s);
UTILS_CHECK_LEN(len = strlen(s));
int32_t count = 0;
for (size_t i = 0; i < len; i ++) {
if (s[i] == letter) {
count ++;
}
}
ret = count * 100 / len;
finish:
return ret;
}
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.