README.md

820. Short Encoding of Words

A valid encoding of an array of words is any reference string s and array of indices indices such that:

• words.length == indices.length
• The reference string s ends with the '#' character.
• For each index indices[i], the substring of s starting from indices[i] and up to (but not including) the next '#' character is equal to words[i].

Given an array of words, return the length of the shortest reference string s possible of any valid encoding of words.

Example 1:

Input: words = ["time", "me", "bell"]
Output: 10
Explanation: A valid encoding would be s = "time#bell#" and indices = [0, 2, 5].
words[0] = "time", the substring of s starting from indices[0] = 0 to the next '#' is underlined in "time#bell#"
words[1] = "me", the substring of s starting from indices[1] = 2 to the next '#' is underlined in "time#bell#"
words[2] = "bell", the substring of s starting from indices[2] = 5 to the next '#' is underlined in "time#bell#"

Example 2:

Input: words = ["t"]
Output: 2
Explanation: A valid encoding would be s = "t#" and indices = [0].

Constraints:

• 1 <= words.length <= 2000
• 1 <= words[i].length <= 7
• words[i] consists of only lowercase letters.

Solutions (Rust)

1. Solution

use std::collections::HashSet;

impl Solution {
    pub fn minimum_length_encoding(words: Vec<String>) -> i32 {
        let words = words.iter().map(|w| w.as_str()).collect::<HashSet<_>>();
        let mut suffixes = HashSet::new();

        for w in &words {
            for i in 1..w.len() {
                suffixes.insert(w.get(i..).unwrap());
            }
        }

        words
            .difference(&suffixes)
            .map(|w| w.len() as i32 + 1)
            .sum()
    }
}