820. Short Encoding of Words

Problem

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.

Solution (Java)

class Solution {
    private static class Node {
        Node[] nodes = new Node[26];
    }

    private boolean insert(Node node, String word) {
        Node current = node;
        int n = word.length();
        boolean flag = false;
        for (int i = n - 1; i >= 0; i--) {
            if (current.nodes[word.charAt(i) - 'a'] == null) {
                current.nodes[word.charAt(i) - 'a'] = new Node();
                flag = true;
            }
            current = current.nodes[word.charAt(i) - 'a'];
        }
        return flag;
    }

    public int minimumLengthEncoding(String[] words) {
        int out = 0;
        Arrays.sort(words, (a, b) -> b.length() - a.length());
        Node node = new Node();
        for (String word : words) {
            if (insert(node, word)) {
                out = out + word.length() + 1;
            }
        }
        return out;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).