2416. Sum of Prefix Scores of Strings

Problem Link

Description

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

We define the score of a string term as the number of strings words[i] such that term 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

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Python3

class TrieNode:
    def __init__(self):
        self.count = 0
        self.children = defaultdict(TrieNode)
 
class Trie:
    def __init__(self):
        self.root = TrieNode()
    
    def insert(self, word):
        curr = self.root
        for x in word:
            curr = curr.children[x]
            curr.count += 1
    
    def query(self, word):
        curr = self.root
        res = 0
 
        for x in word:
            curr = curr.children[x]
            res += curr.count
 
        return res
 
class Solution:
    def sumPrefixScores(self, words: List[str]) -> List[int]:
        trie = Trie()
        res = []
 
        for word in words:
            trie.insert(word)
        
        for word in words:
            res.append(trie.query(word))
 
        return res