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:
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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:
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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
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class Solution {
public:
int minimumLengthEncoding(vector<string>& 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();
}
};