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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.

Solution Explanation:

The problem asks for the shortest reference string that can encode a given array of words. A valid encoding requires that each word is a substring of the reference string, delimited by '#' characters. The solutions leverage a Trie data structure to efficiently find the shortest encoding.

Approach:

Both solutions utilize a Trie, a tree-like data structure used for efficient retrieval of strings. The core idea is to insert the words (reversed) into the Trie. The length of the shortest encoding is related to the number of nodes in the Trie that are leaf nodes (i.e., nodes without children representing the end of a word).

Solution 1: Depth-First Search (DFS) on Trie

  1. Trie Construction: The code first builds a Trie. It iterates through each word in the input array. For each word, it inserts the reversed word into the Trie. Reversing words allows for efficient checking of suffixes.
  2. Depth-First Search (DFS): A DFS function traverses the Trie. Each leaf node represents a unique suffix (in the reversed words). The length of the path from the root to a leaf node plus 1 (for the '#') contributes to the total length of the shortest encoding.
  3. Length Calculation: The dfs function recursively calculates the length contribution of each subtree. The total length of the shortest encoding is the sum of the lengths of paths to all leaf nodes.

Solution 2: Trie Insertion with Length Tracking

This solution is more efficient. It uses a Trie similar to the previous solution. However, instead of doing DFS, it directly tracks the length during insertion:

  1. Sorting: Words are sorted in descending order of their lengths. This is crucial for optimization: longer words are checked first. If a longer word is a suffix of a shorter word, the shorter word will not contribute to the final length because it will already be covered by the longer word.
  2. Trie Insertion and Length Calculation: Each reversed word is inserted into the Trie. The insert function returns the length of the word only if it represents a unique suffix (i.e., it doesn't match a prefix already in the Trie). Otherwise, it returns 0 indicating that it's already been encoded. The sum of these lengths is the final result.

Time Complexity Analysis:

  • Solution 1: The time complexity is dominated by the Trie construction and DFS. Trie construction takes O(Σ|w|) time, where |w| is the length of each word and the summation is over all words. The DFS takes O(Σ|w|) in the worst case (a skewed Trie). Therefore, the overall time complexity is O(Σ|w|).
  • Solution 2: The time complexity is dominated by the Trie insertion. The sorting step takes O(n log n), where n is the number of words. The Trie insertion takes O(Σ|w|) in the worst case. Therefore, the overall time complexity is O(n log n + Σ|w|). Since typically Σ|w| >> n log n, we can approximate it to O(Σ|w|).

Space Complexity Analysis:

For both solutions, the space complexity is O(Σ|w|) due to the Trie structure, which stores all unique suffixes of the words.

In summary, Solution 2 is generally preferred because it avoids the extra overhead of DFS, making it more efficient, especially for a large number of words. Both solutions are correct and provide the same output but with different computational efficiency.