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# 820. Short Encoding of Words

### Description

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`*.*

### Constraints

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

### Approach

### Links

* GeeksforGeeks
* [Leetcode](https://leetcode.com/problems/short-encoding-of-words/)
* ProgramCreek
* YouTube

### **Examples**

{% tabs %}
{% tab title="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#"
{% endtab %}

{% tab title="Example 2" %}
**Input:** words = \["t"]

**Output:** 2

**Explanation:** A valid encoding would be s = "t#" and indices = \[0].
{% endtab %}
{% endtabs %}

### **Solutions**

{% tabs %}
{% tab title="Solution 1" %}

```java
/**
 * Time complexity : O(∑w_i^2), where w_i is the length of words[i].
 * Space complexity : O(∑w_i), the space used in storing suffixes.
 */

class Solution {
    public int minimumLengthEncoding(String[] words) {
        
        Set<String> wordList = new HashSet(Arrays.asList(words));
        
        for(String word: words) {
            for(int j = 1; j < word.length(); j++) {
                wordList.remove(word.substring(j));
            }
        }
        
        int count = 0;
        for(String word: wordList) {
            count += word.length() + 1;
        }
        
        return count;
    }
}
```

{% endtab %}

{% tab title="Solution 2" %}

```java
/**
 * Time complexity : O(∑w_i), where w_i is the length of words[i].
 * Space complexity : O(∑w_i), the space used by the trie.
 */
 
 class Solution {
    public int minimumLengthEncoding(String[] words) {
        TrieNode trie = new TrieNode();
        Map<TrieNode, Integer> nodes = new HashMap();

        for (int i = 0; i < words.length; ++i) {
            String word = words[i];
            TrieNode cur = trie;
            for (int j = word.length() - 1; j >= 0; --j)
                cur = cur.get(word.charAt(j));
            nodes.put(cur, i);
        }

        int ans = 0;
        for (TrieNode node: nodes.keySet()) {
            if (node.count == 0)
                ans += words[nodes.get(node)].length() + 1;
        }
        return ans;

    }
}

class TrieNode {
    TrieNode[] children;
    int count;
    TrieNode() {
        children = new TrieNode[26];
        count = 0;
    }
    public TrieNode get(char c) {
        if (children[c-'a'] == null) {
            children[c-'a'] = new TrieNode();
            count++;
        }
        return children[c - 'a'];
    }
}
```

{% endtab %}
{% endtabs %}

### **Follow up**

*