Description
Given an array of unique strings words, return all the word squares you can build from words. The same word from words can be used multiple times. You can return the answer in any order.
A sequence of strings forms a valid word square if the kth row and column read the same string, where 0 <= k < max(numRows, numColumns).
- For example, the word sequence
["ball","area","lead","lady"] forms a word square because each word reads the same both horizontally and vertically.
Example 1:
Input: words = ["area","lead","wall","lady","ball"]
Output: [["ball","area","lead","lady"],["wall","area","lead","lady"]]
Explanation:
The output consists of two word squares. The order of output does not matter (just the order of words in each word square matters).
Example 2:
Input: words = ["abat","baba","atan","atal"]
Output: [["baba","abat","baba","atal"],["baba","abat","baba","atan"]]
Explanation:
The output consists of two word squares. The order of output does not matter (just the order of words in each word square matters).
Constraints:
1 <= words.length <= 10001 <= words[i].length <= 4- All
words[i] have the same length.
words[i] consists of only lowercase English letters.- All
words[i] are unique.
Solutions
Solution 1
PythonJavaGo
class Trie:
def __init__(self):
self.children = [None] * 26
self.v = []
def insert(self, w, i):
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.v.append(i)
def search(self, w):
node = self
for c in w:
idx = ord(c) - ord('a')
if node.children[idx] is None:
return []
node = node.children[idx]
return node.v
class Solution:
def wordSquares(self, words: List[str]) -> List[List[str]]:
def dfs(t):
if len(t) == len(words[0]):
ans.append(t[:])
return
idx = len(t)
pref = [v[idx] for v in t]
indexes = trie.search(''.join(pref))
for i in indexes:
t.append(words[i])
dfs(t)
t.pop()
trie = Trie()
ans = []
for i, w in enumerate(words):
trie.insert(w, i)
for w in words:
dfs([w])
return ans(code-box)
class Trie {
Trie[] children = new Trie[26];
List<Integer> v = 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.v.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.v;
}
}
class Solution {
private Trie trie = new Trie();
private String[] words;
private List<List<String>> ans = new ArrayList<>();
public List<List<String>> wordSquares(String[] words) {
this.words = words;
for (int i = 0; i < words.length; ++i) {
trie.insert(words[i], i);
}
List<String> t = new ArrayList<>();
for (String w : words) {
t.add(w);
dfs(t);
t.remove(t.size() - 1);
}
return ans;
}
private void dfs(List<String> t) {
if (t.size() == words[0].length()) {
ans.add(new ArrayList<>(t));
return;
}
int idx = t.size();
StringBuilder pref = new StringBuilder();
for (String x : t) {
pref.append(x.charAt(idx));
}
List<Integer> indexes = trie.search(pref.toString());
for (int i : indexes) {
t.add(words[i]);
dfs(t);
t.remove(t.size() - 1);
}
}
}(code-box)
type Trie struct {
children [26]*Trie
v []int
}
func newTrie() *Trie {
return &Trie{}
}
func (this *Trie) insert(word string, i int) {
node := this
for _, c := range word {
c -= 'a'
if node.children[c] == nil {
node.children[c] = newTrie()
}
node = node.children[c]
node.v = append(node.v, i)
}
}
func (this *Trie) search(word string) []int {
node := this
for _, c := range word {
c -= 'a'
if node.children[c] == nil {
return []int{}
}
node = node.children[c]
}
return node.v
}
func wordSquares(words []string) [][]string {
trie := newTrie()
for i, w := range words {
trie.insert(w, i)
}
ans := [][]string{}
var dfs func([]string)
dfs = func(t []string) {
if len(t) == len(words[0]) {
cp := make([]string, len(t))
copy(cp, t)
ans = append(ans, cp)
return
}
idx := len(t)
pref := []byte{}
for _, v := range t {
pref = append(pref, v[idx])
}
indexes := trie.search(string(pref))
for _, i := range indexes {
t = append(t, words[i])
dfs(t)
t = t[:len(t)-1]
}
}
for _, w := range words {
dfs([]string{w})
}
return ans
}(code-box)
