carloscn
commented
11 months ago Analysis
use std::collections::HashSet;
fn has_repeat(input:&Vec<char>) -> bool
{
let mut char_set:HashSet<char> = std::collections::HashSet::new();
for c in input {
if !char_set.insert(*c) {
return true;
}
}
return false;
}
pub fn equal_frequency(word: &str) -> bool
{
if word.is_empty() {
return false;
}
let mut word_vec:Vec<char> = word.chars().collect();
for i in 0..word_vec.len() {
let removed_char: char = word_vec.remove(i);
if !has_repeat(&word_vec) {
return true;
}
word_vec.insert(i, removed_char);
}
return false;
}
Description
You are given a 0-indexed string word, consisting of lowercase English letters. You need to select one index and remove the letter at that index from word so that the frequency of every letter present in word is equal.
Return true if it is possible to remove one letter so that the frequency of all letters in word are equal, and false otherwise.
Note:
The frequency of a letter x is the number of times it occurs in the string. You must remove exactly one letter and cannot choose to do nothing.
Example 1:
Input: word = "abcc" Output: true Explanation: Select index 3 and delete it: word becomes "abc" and each character has a frequency of 1.
Example 2:
Input: word = "aazz" Output: false Explanation: We must delete a character, so either the frequency of "a" is 1 and the frequency of "z" is 2, or vice versa. It is impossible to make all present letters have equal frequency.
Constraints:
2 <= word.length <= 100 word consists of lowercase English letters only.