Description#
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 <= 10001 <= words[i].length <= 1000words[i] consists of lowercase English letters.
Solutions#
Solution 1: Trie#
Use a trie to maintain the prefixes of all strings and the occurrence count of each prefix.
Then, traverse each string, accumulating the occurrence count of each prefix.
The time complexity is $O(n \times m)$. Here, $n$ and $m$ are the length of the string array words and the maximum length of the strings in it, respectively.
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| class Trie:
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]
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| 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;
}
}
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| class Trie {
private:
vector<Trie*> children;
int cnt;
public:
Trie()
: children(26)
, cnt(0) {}
void insert(string& w) {
Trie* node = this;
for (char c : w) {
int idx = c - 'a';
if (!node->children[idx]) node->children[idx] = new Trie();
node = node->children[idx];
++node->cnt;
}
}
int search(string& w) {
Trie* node = this;
int ans = 0;
for (char c : w) {
int idx = c - 'a';
if (!node->children[idx]) return ans;
node = node->children[idx];
ans += node->cnt;
}
return ans;
}
};
class Solution {
public:
vector<int> sumPrefixScores(vector<string>& words) {
Trie* trie = new Trie();
for (auto& w : words) {
trie->insert(w);
}
vector<int> ans;
for (auto& w : words) {
ans.push_back(trie->search(w));
}
return ans;
}
};
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| 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
}
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| function sumPrefixScores(words: string[]): number[] {
const map = new Map();
for (const word of words) {
const n = word.length;
for (let i = 1; i <= n; i++) {
const s = word.slice(0, i);
map.set(s, (map.get(s) ?? 0) + 1);
}
}
return words.map(word => {
const n = word.length;
let count = 0;
for (let i = 1; i <= n; i++) {
const s = word.slice(0, i);
count += map.get(s);
}
return count;
});
}
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Solution 2#
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| 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);
}
let ans = [];
for (const w of words) {
ans.push(trie.search(w));
}
return ans;
}
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