LeetCode #2416 — HARD

Sum of Prefix Scores of Strings

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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].

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Trie

Example 1

["abc","ab","bc","b"]

Example 2

["abcd"]

Core Insight

What unlocks the optimal approach

What data structure will allow you to efficiently keep track of the score of each prefix?
Use a Trie. Insert all the words into it, and keep a counter at each node that will tell you how many times we have visited each prefix.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2416: Sum of Prefix Scores of Strings
class Trie {
    private Trie[] children = new Trie[26];
    private int cnt;

    public void insert(String w) {
        Trie node = this;
        for (char c : w.toCharArray()) {
            c -= 'a';
            if (node.children[c] == null) {
                node.children[c] = new Trie();
            }
            node = node.children[c];
            ++node.cnt;
        }
    }

    public int search(String w) {
        Trie node = this;
        int ans = 0;
        for (char c : w.toCharArray()) {
            c -= 'a';
            if (node.children[c] == null) {
                return ans;
            }
            node = node.children[c];
            ans += node.cnt;
        }
        return ans;
    }
}

class Solution {
    public int[] sumPrefixScores(String[] words) {
        Trie trie = new Trie();
        for (String w : words) {
            trie.insert(w);
        }
        int[] ans = new int[words.length];
        for (int i = 0; i < words.length; ++i) {
            ans[i] = trie.search(words[i]);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2416: Sum of Prefix Scores of Strings
type Trie struct {
	children [26]*Trie
	cnt      int
}

func newTrie() *Trie {
	return &Trie{}
}
func (this *Trie) insert(w string) {
	node := this
	for _, c := range w {
		c -= 'a'
		if node.children[c] == nil {
			node.children[c] = newTrie()
		}
		node = node.children[c]
		node.cnt++
	}
}

func (this *Trie) search(word string) int {
	node := this
	ans := 0
	for _, c := range word {
		c -= 'a'
		if node.children[c] == nil {
			return ans
		}
		node = node.children[c]
		ans += node.cnt
	}
	return ans
}

func sumPrefixScores(words []string) []int {
	trie := newTrie()
	for _, w := range words {
		trie.insert(w)
	}
	ans := make([]int, len(words))
	for i, w := range words {
		ans[i] = trie.search(w)
	}
	return ans
}

# Accepted solution for LeetCode #2416: Sum of Prefix Scores of Strings
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]

// Accepted solution for LeetCode #2416: Sum of Prefix Scores of Strings
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2416: Sum of Prefix Scores of Strings
// class Trie {
//     private Trie[] children = new Trie[26];
//     private int cnt;
// 
//     public void insert(String w) {
//         Trie node = this;
//         for (char c : w.toCharArray()) {
//             c -= 'a';
//             if (node.children[c] == null) {
//                 node.children[c] = new Trie();
//             }
//             node = node.children[c];
//             ++node.cnt;
//         }
//     }
// 
//     public int search(String w) {
//         Trie node = this;
//         int ans = 0;
//         for (char c : w.toCharArray()) {
//             c -= 'a';
//             if (node.children[c] == null) {
//                 return ans;
//             }
//             node = node.children[c];
//             ans += node.cnt;
//         }
//         return ans;
//     }
// }
// 
// class Solution {
//     public int[] sumPrefixScores(String[] words) {
//         Trie trie = new Trie();
//         for (String w : words) {
//             trie.insert(w);
//         }
//         int[] ans = new int[words.length];
//         for (int i = 0; i < words.length; ++i) {
//             ans[i] = trie.search(words[i]);
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2416: Sum of Prefix Scores of Strings
class Trie {
    children: Array<any>;
    cnt: number;

    constructor() {
        this.children = Array(26);
        this.cnt = 0;
    }

    insert(w: string): void {
        let node = this;
        for (const c of w) {
            const idx = c.charCodeAt(0) - 'a'.charCodeAt(0);
            if (!node.children[idx]) {
                node.children[idx] = new Trie();
            }
            node = node.children[idx];
            node.cnt++;
        }
    }

    search(w: string): number {
        let node = this;
        let ans = 0;
        for (const c of w) {
            const idx = c.charCodeAt(0) - 'a'.charCodeAt(0);
            if (!node.children[idx]) {
                return ans;
            }
            node = node.children[idx];
            ans += node.cnt;
        }
        return ans;
    }
}

function sumPrefixScores(words: string[]): number[] {
    const trie = new Trie();
    for (const w of words) {
        trie.insert(w);
    }
    return words.map(w => trie.search(w));
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(L)

Space

O(L)

Approach Breakdown

HASH SET

O(N × L) time

O(N × L) space

Store all N words in a hash set. Each insert/lookup hashes the entire word of length L, giving O(L) per operation. Prefix queries require checking every stored word against the prefix — O(N × L) per prefix search. Space is O(N × L) for storing all characters.

TRIE

O(L) time

O(N × L) space

Each operation (insert, search, prefix) takes O(L) time where L is the word length — one node visited per character. Total space is bounded by the sum of all stored word lengths. Tries win over hash sets when you need prefix matching: O(L) prefix search vs. checking every stored word.

Shortcut: One node per character → O(L) per operation. Prefix queries are what make tries worthwhile.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.