2416. Sum of Prefix Scores of Strings

Description

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.

 

Solutions

Solution: Trie

• Time complexity: O(n*Max(word.length))
• Space complexity: O(n*Max(word.length))

 

JavaScript

/**
 * @param {string[]} words
 * @return {number[]}
 */
const sumPrefixScores = function (words) {
  const trie = {};

  for (const word of words) {
    let node = trie;

    for (const letter of word) {
      if (!node[letter]) {
        node[letter] = { count: 0 };
      }
      node = node[letter];
      node.count += 1;
    }
  }

  return words.map(word => {
    let node = trie;
    let scores = 0;

    for (const letter of word) {
      node = node[letter];
      scores += node.count;
    }
    return scores;
  });
};