carloscn
commented
1 year ago Analysis
pub fn are_occurrences_equal(s: &str) -> bool
{
if s.len() < 1 {
return false;
}
let mut s_vec:Vec<char> = s.chars().collect();
let mut k_vec:Vec<char> = vec![];
let mut count:i32;
let mut last_count:i32 = 0;
let mut first = true;
s_vec.sort();
s_vec.push(' ');
for i in 0..s_vec.len() {
if k_vec.is_empty() {
k_vec.push(s_vec[i]);
continue;
} else {
if k_vec[k_vec.len() - 1] != s_vec[i] {
count = k_vec.len() as i32;
if last_count != count {
if first == false {
return false;
}
}
last_count = count;
first = false;
k_vec.clear();
}
k_vec.push(s_vec[i]);
}
}
return true;
}
Description
Given a string s, return true if s is a good string, or false otherwise.
A string s is good if all the characters that appear in s have the same number of occurrences (i.e., the same frequency).
Example 1:
Input: s = "abacbc" Output: true Explanation: The characters that appear in s are 'a', 'b', and 'c'. All characters occur 2 times in s.
Example 2:
Input: s = "aaabb" Output: false Explanation: The characters that appear in s are 'a' and 'b'. 'a' occurs 3 times while 'b' occurs 2 times, which is not the same number of times.
Constraints:
1 <= s.length <= 1000 s consists of lowercase English letters.