## 748. Shortest Completing Word

Given a string `licensePlate` and an array of strings `words`, find the shortest completing word in `words`.

A completing word is a word that contains all the letters in `licensePlate`. Ignore numbers and spaces in `licensePlate`, and treat letters as case insensitive. If a letter appears more than once in `licensePlate`, then it must appear in the word the same number of times or more.

For example, if `licensePlate`` = "aBc 12c"`, then it contains letters `'a'`, `'b'` (ignoring case), and `'c'` twice. Possible completing words are `"abccdef"`, `"caaacab"`, and `"cbca"`.

Return the shortest completing word in `words`. It is guaranteed an answer exists. If there are multiple shortest completing words, return the first one that occurs in `words`.

Example 1:

```Input: licensePlate = "1s3 PSt", words = ["step","steps","stripe","stepple"]
Output: "steps"
Explanation: licensePlate contains letters 's', 'p', 's' (ignoring case), and 't'.
"step" contains 't' and 'p', but only contains 1 's'.
"steps" contains 't', 'p', and both 's' characters.
"stripe" is missing an 's'.
"stepple" is missing an 's'.
Since "steps" is the only word containing all the letters, that is the answer.
```

Example 2:

```Input: licensePlate = "1s3 456", words = ["looks","pest","stew","show"]
Output: "pest"
Explanation: licensePlate only contains the letter 's'. All the words contain 's', but among these "pest", "stew", and "show" are shortest. The answer is "pest" because it is the word that appears earliest of the 3.
```

Example 3:

```Input: licensePlate = "Ah71752", words = ["suggest","letter","of","husband","easy","education","drug","prevent","writer","old"]
Output: "husband"
```

Example 4:

```Input: licensePlate = "OgEu755", words = ["enough","these","play","wide","wonder","box","arrive","money","tax","thus"]
Output: "enough"
```

Example 5:

```Input: licensePlate = "iMSlpe4", words = ["claim","consumer","student","camera","public","never","wonder","simple","thought","use"]
Output: "simple"
```

Constraints:

• `1 <= licensePlate.length <= 7`
• `licensePlate` contains digits, letters (uppercase or lowercase), or space `' '`.
• `1 <= words.length <= 1000`
• `1 <= words[i].length <= 15`
• `words[i]` consists of lower case English letters.

## Rust Solution

``````struct Solution;

#[derive(Debug, PartialEq, Eq, Clone)]
struct Count {
v: Vec<i32>,
}

impl Count {
fn new(s: &str) -> Self {
let mut v: Vec<i32> = vec![0; 256];
for c in s.chars() {
v[c as usize] += 1;
}
Count { v }
}

fn completes(&self, other: &Count) -> bool {
for i in 0..26 {
let c: usize = (b'a' + i) as usize;
if self.v[c] < other.v[c] {
return false;
}
}
true
}
}

impl Solution {
fn shortest_completing_word(license_plate: String, words: Vec<String>) -> String {
let mut min: Option<String> = None;
.chars()
.filter(|c| c.is_alphabetic())
.collect();
let lowercase = letters.to_lowercase();
let count_of_lowercase = Count::new(&lowercase);
for word in words {
let count_of_word = Count::new(&word);
if count_of_word.completes(&count_of_lowercase) {
if let Some(ref s) = min {
if word.len() < s.len() {
min = Some(word);
}
} else {
min = Some(word);
}
}
}
min.unwrap()
}
}

#[test]
fn test() {
let words: Vec<String> = vec_string!["step", "steps", "stripe", "stepple"];
let res = "steps".to_string();
assert_eq!(