# Short Encoding Of Words Problem

## Description

LeetCode Problem 820.

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:

 1 2 3 4 5 6 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:

 1 2 3 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.

## Sample C++ Code

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 class Solution { public: int minimumLengthEncoding(vector& words) { sort(words.begin(), words.end(), sortByLen); string encoded = words[0] + "#"; for (int i = 1; i < words.size(); i++) { if (encoded.find(words[i] + "#") == -1) encoded += words[i] + "#"; } return encoded.size(); } private: static bool sortByLen(string &a, string &b) { return a.size() > b.size(); } };