You are given an array words of size n consisting of non-empty strings.

We define the score of a string word as the number of strings words[i] such that word is a prefix of words[i].

• For example, if words = ["a", "ab", "abc", "cab"], then the score of "ab" is 2, since "ab" is a prefix of both "ab" and "abc".

Return an array answer of size n where answer[i] is the sum of scores of every non-empty prefix of words[i].

Note that a string is considered as a prefix of itself.

Example 1:

Input: words = ["abc","ab","bc","b"]
Output: [5,4,3,2]
Explanation: The answer for each string is the following:
- "abc" has 3 prefixes: "a", "ab", and "abc".
- There are 2 strings with the prefix "a", 2 strings with the prefix "ab", and 1 string with the prefix "abc".
The total is answer[0] = 2 + 2 + 1 = 5.
- "ab" has 2 prefixes: "a" and "ab".
- There are 2 strings with the prefix "a", and 2 strings with the prefix "ab".
The total is answer[1] = 2 + 2 = 4.
- "bc" has 2 prefixes: "b" and "bc".
- There are 2 strings with the prefix "b", and 1 string with the prefix "bc".
The total is answer[2] = 2 + 1 = 3.
- "b" has 1 prefix: "b".
- There are 2 strings with the prefix "b".
The total is answer[3] = 2.

Example 2:

Input: words = ["abcd"]
Output: [4]
Explanation:
"abcd" has 4 prefixes: "a", "ab", "abc", and "abcd".
Each prefix has a score of one, so the total is answer[0] = 1 + 1 + 1 + 1 = 4.

Constraints:

• 1 <= words.length <= 1000
• 1 <= words[i].length <= 1000
• words[i] consists of lowercase English letters.

Solution: Trie

Insert all the words into a tire whose node val is the number of substrings that have the current prefix.

During query time, sum up the values along the prefix path.

Time complexity: O(sum(len(word))
Space complexity: O(sum(len(word))

// Author: Huahua
struct Trie {
  Trie* ch[26] = {};
  int cnt = 0;
  void insert(string_view s) {
    Trie* cur = this;
    for (char c : s) {
      if (!cur->ch[c - 'a']) 
        cur->ch[c - 'a'] = new Trie();
      cur = cur->ch[c - 'a'];
      ++cur->cnt;
    }
  }
  int query(string_view s) {
    Trie* cur = this;
    int ans = 0;
    for (char c : s) {
      cur = cur->ch[c - 'a'];
      ans += cur->cnt;
    }
    return ans;
  }
};
class Solution {
public:
  vector<int> sumPrefixScores(vector<string>& words) {
    Trie* root = new Trie();
    for (const string& w : words)
      root->insert(w);
    vector<int> ans;
    for (const string& w : words)
      ans.push_back(root->query(w));
    return ans;
  }
};

The post 花花酱 LeetCode 2416. Sum of Prefix Scores of Strings appeared first on Huahua's Tech Road.