A Trie is an efficient tree-like data structure that allows efficient retrieval of a string's prefix scores. We start by inserting each word into the Trie, along with maintaining a count at each node to track how many words share that prefix. Then, for each word, we traverse its prefixes in the Trie to compute the prefix score by collecting the counts stored at each relevant node.
The time complexity is O(n * m) where n is the number of words and m is the maximum length of the words, due to Trie insertions and lookups. The space complexity is O(N * m), N being the total length of all words combined, as each character might lead to a node in the Trie.
1class TrieNode {
2 constructor() {
3 this.children = {};
4 this.count = 0;
5 }
6}
7
8class Trie {
9 constructor() {
10 this.root = new TrieNode();
11 }
12
13 insert(word) {
14 let node = this.root;
15 for (let char of word) {
16 if (!node.children[char]) {
17 node.children[char] = new TrieNode();
18 }
19 node = node.children[char];
20 node.count++;
21 }
22 }
23
24 calculatePrefixScore(word) {
25 let node = this.root;
26 let score = 0;
27 for (let char of word) {
28 node = node.children[char];
29 score += node.count;
30 }
31 return score;
32 }
33}
34
35function sumOfPrefixScores(words) {
36 const trie = new Trie();
37 for (let word of words) {
38 trie.insert(word);
39 }
40 return words.map(word => trie.calculatePrefixScore(word));
41}
42
43const words = ["abc", "ab", "bc", "b"];
44console.log(sumOfPrefixScores(words));The JavaScript solution constructs a Trie for the input words, incrementing counters at relevant nodes as each word is inserted. For prefix score calculation, it traverses through the Trie nodes relevant to each word, summing up their maintained counts to form the final result.
We can also use a hashtable (or dictionary) to count how often each prefix appears in the list of words. Iterate through each word and generate all its prefixes, updating their counts in the hashtable. Afterwards, compute the sum of counts for each full word by referencing its prefixes in the hashtable.
Time complexity is O(n * m) as we process each prefix per word insertion and retrieval. Space complexity is potentially O(N * m), accounting for hash table storage of prefixes.
1#include <stdio.h>
2#include <stdlib.h>
3#include <string.h>
4
5#define MAX_WORD_LENGTH 1000
6#define TABLE_SIZE 1000003
7
8typedef struct Node {
9 char *key;
10 int value;
11 struct Node *next;
12} Node;
13
14unsigned long hash(const char *str) {
15 unsigned long hash = 5381;
16 int c;
17 while ((c = *str++))
18 hash = ((hash << 5) + hash) + c;
19 return hash;
20}
21
22void insert(Node **table, const char *key) {
23 unsigned long index = hash(key) % TABLE_SIZE;
24 Node *entry = table[index];
25 while (entry != NULL) {
26 if (strcmp(entry->key, key) == 0) {
27 entry->value += 1;
28 return;
29 }
30 entry = entry->next;
31 }
32 Node *new_node = malloc(sizeof(Node));
33 new_node->key = strdup(key);
34 new_node->value = 1;
35 new_node->next = table[index];
36 table[index] = new_node;
37}
38
39int get(Node **table, const char *key) {
40 unsigned long index = hash(key) % TABLE_SIZE;
41 Node *entry = table[index];
42 while (entry != NULL) {
43 if (strcmp(entry->key, key) == 0)
44 return entry->value;
45 entry = entry->next;
46 }
47 return 0;
48}
49
50void prefixScores(char *words[], int n, int *answer) {
51 Node *table[TABLE_SIZE] = {NULL};
52 char prefix[MAX_WORD_LENGTH];
53 for (int i = 0; i < n; i++) {
54 strcpy(prefix, "");
55 for (int j = 0; words[i][j] != '\0'; j++) {
56 prefix[j] = words[i][j];
57 prefix[j + 1] = '\0';
58 insert(table, prefix);
59 }
60 }
61 for (int i = 0; i < n; i++) {
62 strcpy(prefix, "");
63 answer[i] = 0;
64 for (int j = 0; words[i][j] != '\0'; j++) {
65 prefix[j] = words[i][j];
66 prefix[j + 1] = '\0';
67 answer[i] += get(table, prefix);
68 }
69 }
70}
71
72int main() {
73 char *words[] = {"abc", "ab", "bc", "b"};
74 int n = sizeof(words) / sizeof(words[0]);
75 int answer[n];
76 prefixScores(words, n, answer);
77 for (int i = 0; i < n; i++)
78 printf("answer[%d] = %d\n", i, answer[i]);
79 return 0;
80}The C solution uses a hash table to keep track of prefix counts. For every word, we insert each of its prefixes into the hash table, updating their counts. Then, by iterating over prefixes of each word and retrieving their counts from the hash table, it calculates the prefix scores.