Leetcode #2416: Sum of Prefix Scores of Strings
In this guide, we solve Leetcode #2416 Sum of Prefix Scores of Strings in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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].
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Trie, Array, String, Counting
Intuition
Prefix queries are most efficient with a trie.
Each character transitions to the next node in the tree.
Approach
Insert words into the trie and traverse by characters for queries.
Track terminal markers to distinguish full words from prefixes.
Steps:
- Build the trie.
- Traverse for each query.
- Return matches or validations.
Example
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.
Python Solution
class Trie:
__slots__ = "children", "cnt"
def __init__(self):
self.children = [None] * 26
self.cnt = 0
def insert(self, w):
node = self
for c in w:
idx = ord(c) - ord("a")
if node.children[idx] is None:
node.children[idx] = Trie()
node = node.children[idx]
node.cnt += 1
def search(self, w):
node = self
ans = 0
for c in w:
idx = ord(c) - ord("a")
if node.children[idx] is None:
return ans
node = node.children[idx]
ans += node.cnt
return ans
class Solution:
def sumPrefixScores(self, words: List[str]) -> List[int]:
trie = Trie()
for w in words:
trie.insert(w)
return [trie.search(w) for w in words]
Complexity
The time complexity is O(total characters). The space complexity is O(total characters).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.