2416. Sum of Prefix Scores of Strings

Problem

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 (Java)

class Trie {
    Trie[] ch = new Trie[26];
    int visited = 0;
}
class Solution {
    public int[] sumPrefixScores(String[] words) {
        Trie trie = new Trie();
        int[] ans = new int[words.length];
        int k = 0;
        for (String x: words) {
            Trie t = trie;
            for (int i = 0; i < x.length(); i++) {
                int c = x.charAt(i) - 'a';
                if (t.ch[c] == null) t.ch[c] = new Trie();
                t.ch[c].visited++;
                t = t.ch[c];
            }
        }
        for (String x: words) {
            Trie t = trie; int curr = 0;
            for (int i = 0; i < x.length(); i++) {
                int c = x.charAt(i) - 'a';
                curr += t.ch[c].visited;
                t = t.ch[c];
            }
            ans[k++] = curr;
        }
        return ans;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).