Given a string band an array of smaller strings T, design a method to search b for each small string in T. Output positions
of all strings in smalls
that appear in big
, where positions[i]
is all positions of smalls[i]
.
Example:
Input: big = "mississippi" smalls = ["is","ppi","hi","sis","i","ssippi"] Output: [[1,4],[8],[],[3],[1,4,7,10],[5]]
Note:
0 <= len(big) <= 1000
0 <= len(smalls[i]) <= 1000
- The total number of characters in
smalls
will not exceed 100000. - No duplicated strings in
smalls
. - All characters are lowercase letters.
class Trie:
def __init__(self):
self.idx = -1
self.children = [None] * 26
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.idx = i
def search(self, word):
res = []
node = self
for c in word:
idx = ord(c) - ord('a')
if node.children[idx] is None:
return res
node = node.children[idx]
if node.idx != -1:
res.append(node.idx)
return res
class Solution:
def multiSearch(self, big: str, smalls: List[str]) -> List[List[int]]:
tree = Trie()
for i, s in enumerate(smalls):
tree.insert(s, i)
n = len(smalls)
ans = [[] for _ in range(n)]
for i in range(len(big)):
s = big[i:]
for idx in tree.search(s):
ans[idx].append(i)
return ans
class Solution {
public int[][] multiSearch(String big, String[] smalls) {
Trie tree = new Trie();
int n = smalls.length;
for (int i = 0; i < n; ++i) {
tree.insert(smalls[i], i);
}
List<List<Integer>> res = new ArrayList<>();
for (int i = 0; i < n; ++i) {
res.add(new ArrayList<>());
}
for (int i = 0; i < big.length(); ++i) {
String s = big.substring(i);
List<Integer> t = tree.search(s);
for (int idx : t) {
res.get(idx).add(i);
}
}
int[][] ans = new int[n][];
for (int i = 0; i < n; ++i) {
ans[i] = res.get(i).stream().mapToInt(Integer::intValue).toArray();
}
return ans;
}
}
class Trie {
private int idx;
private Trie[] children;
public Trie() {
idx = -1;
children = new Trie[26];
}
public 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.idx = i;
}
public List<Integer> search(String word) {
Trie node = this;
List<Integer> res = new ArrayList<>();
for (char c : word.toCharArray()) {
c -= 'a';
if (node.children[c] == null) {
return res;
}
node = node.children[c];
if (node.idx != -1) {
res.add(node.idx);
}
}
return res;
}
}
class Trie {
private:
vector<Trie*> children;
int idx;
public:
Trie()
: children(26)
, idx(-1) { }
void insert(string word, int i) {
Trie* node = this;
for (char c : word) {
int idx = c - 'a';
if (!node->children[idx]) node->children[idx] = new Trie();
node = node->children[idx];
}
node->idx = i;
}
vector<int> search(string word) {
Trie* node = this;
vector<int> res;
for (char c : word) {
int idx = c - 'a';
if (!node->children[idx]) return res;
node = node->children[idx];
if (node->idx != -1) res.push_back(node->idx);
}
return res;
}
};
class Solution {
public:
vector<vector<int>> multiSearch(string big, vector<string>& smalls) {
Trie* tree = new Trie();
int n = smalls.size();
for (int i = 0; i < n; ++i) tree->insert(smalls[i], i);
vector<vector<int>> ans(n);
for (int i = 0, m = big.size(); i < m; ++i) {
string s = big.substr(i, m - i);
vector<int> t = tree->search(s);
for (int& idx : t) ans[idx].push_back(i);
}
return ans;
}
};
type Trie struct {
children [26]*Trie
idx int
}
func newTrie() *Trie {
return &Trie{idx: -1}
}
func (this *Trie) insert(word string, idx int) {
node := this
for _, c := range word {
idx := c - 'a'
if node.children[idx] == nil {
node.children[idx] = newTrie()
}
node = node.children[idx]
}
node.idx = idx
}
func (this *Trie) search(word string) []int {
node := this
var res []int
for _, c := range word {
idx := c - 'a'
if node.children[idx] == nil {
return res
}
node = node.children[idx]
if node.idx != -1 {
res = append(res, node.idx)
}
}
return res
}
func multiSearch(big string, smalls []string) [][]int {
tree := newTrie()
for i, s := range smalls {
tree.insert(s, i)
}
n := len(smalls)
ans := make([][]int, n)
for i := range big {
s := big[i:]
t := tree.search(s)
for _, idx := range t {
ans[idx] = append(ans[idx], i)
}
}
return ans
}