How do you solve Sum of Prefix Scores of Strings on LeetCode?

Sum of Prefix Scores of Strings (LeetCode #2416) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(n * m) space complexity. Key insight: Using a Trie allows efficient prefix counting as it organizes words hierarchically.

What is the brute force solution for Sum of Prefix Scores of Strings?

The brute force approach for Sum of Prefix Scores of Strings is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking each prefix of every word against all words in the list to count how many words start with that prefix. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n * m).

What algorithm or data structure is used to solve Sum of Prefix Scores of Strings?

Sum of Prefix Scores of Strings uses the following concepts: Array, String, Trie, Counting. The recommended patterns to study are: Trie, Prefix Tree, Hash Map.

What are the interview tips for Sum of Prefix Scores of Strings?

Always think about how you can optimize a brute-force solution — interviewers appreciate candidates who can identify inefficiencies. Be ready to explain your data structure choice clearly, especially if you opt for something like a Trie. Practice explaining your thought process out loud as you code; it helps interviewers follow your logic.

What are common mistakes when solving Sum of Prefix Scores of Strings?

Not considering the efficiency of prefix checking, leading to unnecessary repeated calculations. Overlooking the need to handle edge cases like empty strings or single-character words.

#2416

Sum of Prefix Scores of Strings

Hard

ArrayStringTrieCountingTriePrefix TreeHash Map

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m)
Space	O(1)	O(n * m)

💡

Intuition

Time O(n * m)Space O(n * m)

Using a Trie allows us to efficiently count the occurrences of prefixes as we insert words. Each node in the Trie can keep track of how many words pass through it, making it easy to compute scores for each prefix.

⚙️

Algorithm

3 steps

1Step 1: Create a Trie data structure and insert all words into it, updating the count at each node.
2Step 2: For each word, traverse its prefixes in the Trie and sum the counts stored at each node.
3Step 3: Store the total score for each word in the result array.

solution.py31 lines

1class TrieNode:
2    def __init__(self):
3        self.children = {}
4        self.count = 0
5
6class Trie:
7    def __init__(self):
8        self.root = TrieNode()
9
10    def insert(self, word):
11        node = self.root
12        for char in word:
13            if char not in node.children:
14                node.children[char] = TrieNode()
15            node = node.children[char]
16            node.count += 1
17
18    def get_score(self, word):
19        node = self.root
20        score = 0
21        for char in word:
22            node = node.children[char]
23            score += node.count
24        return score
25
26
27def sumPrefixScores(words):
28    trie = Trie()
29    for word in words:
30        trie.insert(word)
31    return [trie.get_score(word) for word in words]

ℹ

Complexity note: The time complexity is O(n * m) where n is the number of words and m is the average length of the words. The space complexity is O(n * m) due to the Trie storing all characters of the words.

1Using a Trie allows efficient prefix counting as it organizes words hierarchically.
2Each node in the Trie can maintain a count of how many words pass through it, making score calculation straightforward.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.