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

• ProgramCreek

• YouTube

Examples

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].

Solutions

Solution 1

/**
 * 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;
    }
}

Solution 2

/**
 * 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'];
    }
}

Follow up