Description#
Design a special dictionary that searches the words in it by a prefix and a suffix.
Implement the WordFilter class:
WordFilter(string[] words) Initializes the object with the words in the dictionary.f(string pref, string suff) Returns the index of the word in the dictionary, which has the prefix pref and the suffix suff. If there is more than one valid index, return the largest of them. If there is no such word in the dictionary, return -1.
Example 1:
Input
["WordFilter", "f"]
[[["apple"]], ["a", "e"]]
Output
[null, 0]
Explanation
WordFilter wordFilter = new WordFilter(["apple"]);
wordFilter.f("a", "e"); // return 0, because the word at index 0 has prefix = "a" and suffix = "e".
Constraints:
1 <= words.length <= 1041 <= words[i].length <= 71 <= pref.length, suff.length <= 7words[i], pref and suff consist of lowercase English letters only.- At most
104 calls will be made to the function f.
Solutions#
Solution 1#
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| class WordFilter:
def __init__(self, words: List[str]):
self.d = {}
for k, w in enumerate(words):
n = len(w)
for i in range(n + 1):
a = w[:i]
for j in range(n + 1):
b = w[j:]
self.d[(a, b)] = k
def f(self, pref: str, suff: str) -> int:
return self.d.get((pref, suff), -1)
# Your WordFilter object will be instantiated and called as such:
# obj = WordFilter(words)
# param_1 = obj.f(pref,suff)
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| class WordFilter {
private Map<String, Integer> d = new HashMap<>();
public WordFilter(String[] words) {
for (int k = 0; k < words.length; ++k) {
String w = words[k];
int n = w.length();
for (int i = 0; i <= n; ++i) {
String a = w.substring(0, i);
for (int j = 0; j <= n; ++j) {
String b = w.substring(j);
d.put(a + "." + b, k);
}
}
}
}
public int f(String pref, String suff) {
return d.getOrDefault(pref + "." + suff, -1);
}
}
/**
* Your WordFilter object will be instantiated and called as such:
* WordFilter obj = new WordFilter(words);
* int param_1 = obj.f(pref,suff);
*/
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| class WordFilter {
public:
unordered_map<string, int> d;
WordFilter(vector<string>& words) {
for (int k = 0; k < words.size(); ++k) {
string w = words[k];
int n = w.size();
for (int i = 0; i <= n; ++i) {
string a = w.substr(0, i);
for (int j = 0; j <= n; ++j) {
string b = w.substr(j, n - j);
d[a + "." + b] = k;
}
}
}
}
int f(string pref, string suff) {
string key = pref + "." + suff;
if (d.count(key)) return d[key];
return -1;
}
};
/**
* Your WordFilter object will be instantiated and called as such:
* WordFilter* obj = new WordFilter(words);
* int param_1 = obj->f(pref,suff);
*/
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| type WordFilter struct {
d map[string]int
}
func Constructor(words []string) WordFilter {
d := map[string]int{}
for k, w := range words {
n := len(w)
for i := 0; i <= n; i++ {
a := w[:i]
for j := 0; j <= n; j++ {
b := w[j:]
d[a+"."+b] = k
}
}
}
return WordFilter{d}
}
func (this *WordFilter) F(pref string, suff string) int {
if v, ok := this.d[pref+"."+suff]; ok {
return v
}
return -1
}
/**
* Your WordFilter object will be instantiated and called as such:
* obj := Constructor(words);
* param_1 := obj.F(pref,suff);
*/
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Solution 2#
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| class Trie:
def __init__(self):
self.children = [None] * 26
self.indexes = []
def insert(self, word, i):
node = self
for c in word:
idx = ord(c) - ord("a")
if node.children[idx] is None:
node.children[idx] = Trie()
node = node.children[idx]
node.indexes.append(i)
def search(self, pref):
node = self
for c in pref:
idx = ord(c) - ord("a")
if node.children[idx] is None:
return []
node = node.children[idx]
return node.indexes
class WordFilter:
def __init__(self, words: List[str]):
self.p = Trie()
self.s = Trie()
for i, w in enumerate(words):
self.p.insert(w, i)
self.s.insert(w[::-1], i)
def f(self, pref: str, suff: str) -> int:
a = self.p.search(pref)
b = self.s.search(suff[::-1])
if not a or not b:
return -1
i, j = len(a) - 1, len(b) - 1
while ~i and ~j:
if a[i] == b[j]:
return a[i]
if a[i] > b[j]:
i -= 1
else:
j -= 1
return -1
# Your WordFilter object will be instantiated and called as such:
# obj = WordFilter(words)
# param_1 = obj.f(pref,suff)
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| class Trie {
Trie[] children = new Trie[26];
List<Integer> indexes = new ArrayList<>();
void insert(String word, int i) {
Trie node = this;
for (char c : word.toCharArray()) {
c -= 'a';
if (node.children[c] == null) {
node.children[c] = new Trie();
}
node = node.children[c];
node.indexes.add(i);
}
}
List<Integer> search(String pref) {
Trie node = this;
for (char c : pref.toCharArray()) {
c -= 'a';
if (node.children[c] == null) {
return Collections.emptyList();
}
node = node.children[c];
}
return node.indexes;
}
}
class WordFilter {
private Trie p = new Trie();
private Trie s = new Trie();
public WordFilter(String[] words) {
for (int i = 0; i < words.length; ++i) {
String w = words[i];
p.insert(w, i);
s.insert(new StringBuilder(w).reverse().toString(), i);
}
}
public int f(String pref, String suff) {
suff = new StringBuilder(suff).reverse().toString();
List<Integer> a = p.search(pref);
List<Integer> b = s.search(suff);
if (a.isEmpty() || b.isEmpty()) {
return -1;
}
int i = a.size() - 1, j = b.size() - 1;
while (i >= 0 && j >= 0) {
int c = a.get(i), d = b.get(j);
if (c == d) {
return c;
}
if (c > d) {
--i;
} else {
--j;
}
}
return -1;
}
}
/**
* Your WordFilter object will be instantiated and called as such:
* WordFilter obj = new WordFilter(words);
* int param_1 = obj.f(pref,suff);
*/
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| type Trie struct {
children [26]*Trie
indexes []int
}
func newTrie() *Trie {
return &Trie{indexes: []int{}}
}
func (this *Trie) insert(word string, i int) {
node := this
for _, c := range word {
idx := c - 'a'
if node.children[idx] == nil {
node.children[idx] = newTrie()
}
node = node.children[idx]
node.indexes = append(node.indexes, i)
}
}
func (this *Trie) search(pref string) []int {
node := this
for _, c := range pref {
idx := c - 'a'
if node.children[idx] == nil {
return []int{}
}
node = node.children[idx]
}
return node.indexes
}
type WordFilter struct {
p *Trie
s *Trie
}
func Constructor(words []string) WordFilter {
p := newTrie()
s := newTrie()
for i, w := range words {
p.insert(w, i)
s.insert(reverse(w), i)
}
return WordFilter{p, s}
}
func (this *WordFilter) F(pref string, suff string) int {
a := this.p.search(pref)
b := this.s.search(reverse(suff))
if len(a) == 0 || len(b) == 0 {
return -1
}
i, j := len(a)-1, len(b)-1
for i >= 0 && j >= 0 {
if a[i] == b[j] {
return a[i]
}
if a[i] > b[j] {
i--
} else {
j--
}
}
return -1
}
func reverse(w string) string {
ww := []byte(w)
for i, j := 0, len(w)-1; i < j; i++ {
ww[i], ww[j] = ww[j], ww[i]
j--
}
return string(ww)
}
/**
* Your WordFilter object will be instantiated and called as such:
* obj := Constructor(words);
* param_1 := obj.F(pref,suff);
*/
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