2494. Sum Of Prefix Scores Of Strings¶

Difficulty: Hard

2494. Sum of Prefix Scores of Strings

Hard

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¶

class TrieNode {
    public int freq;
    public TrieNode[] children;

    public TrieNode() {
        freq = 0;
        children = new TrieNode[26];
    }
}

class Solution {
    public int[] sumPrefixScores(String[] words) {
        int n = words.length;
        int[] ans = new int[n];
        TrieNode root = new TrieNode();
        for(String w: words) {
            TrieNode curr = root;
            for(char c: w.toCharArray()) {
                if(curr.children[c - 'a'] == null)
                    curr.children[c - 'a'] = new TrieNode();
                curr = curr.children[c - 'a'];
                curr.freq++;
            }
        }
        for(int i = 0; i < n; i++) {
            TrieNode curr = root;
            for(char c: words[i].toCharArray()) {
                curr = curr.children[c - 'a'];
                ans[i] += curr.freq;
            }
        }
        return ans;
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here