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true |
Hard |
1725 |
Weekly Contest 311 Q4 |
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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"is2, 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.
We can use a prefix tree to maintain all prefixes of the strings and count the occurrences of each prefix.
Define the prefix tree node structure Trie, which includes two properties:
children: An array of length 26 used to store the current node's children.cnt: The occurrence count of the current node.
Define two methods for the prefix tree:
insert: Inserts a string, adding its prefixes into the prefix tree.search: Searches for a string and returns the occurrence count of its prefixes.
We traverse all strings, inserting each string into the prefix tree. Then we traverse all strings again, calling the search method for each string and summing up the occurrence counts of each prefix.
Time complexity is
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]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;
}
}class Trie {
private:
Trie* children[26]{};
int cnt = 0;
public:
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;
}
};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
}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));
}class Trie {
constructor() {
this.children = {};
this.cnt = 0;
}
insert(w) {
let node = this;
for (const c of w) {
if (!node.children[c]) {
node.children[c] = new Trie();
}
node = node.children[c];
node.cnt++;
}
}
search(w) {
let node = this;
let ans = 0;
for (const c of w) {
if (!node.children[c]) {
return ans;
}
node = node.children[c];
ans += node.cnt;
}
return ans;
}
}
/**
* @param {string[]} words
* @return {number[]}
*/
var sumPrefixScores = function (words) {
const trie = new Trie();
for (const w of words) {
trie.insert(w);
}
return words.map(w => trie.search(w));
};