2.3 KiB
2.3 KiB
TODO 1408. String Matching in an Array easy
Given an array of string words, return all strings in/ words /that are a substring of another word. You can return the answer in any order.
Example 1:
Input: words = ["mass","as","hero","superhero"]
Output: ["as","hero"]
Explanation: "as" is substring of "mass" and "hero" is substring of "superhero".
["hero","as"] is also a valid answer.
Example 2:
Input: words = ["leetcode","et","code"]
Output: ["et","code"]
Explanation: "et", "code" are substring of "leetcode".
Example 3:
Input: words = ["blue","green","bu"]
Output: []
Explanation: No string of words is substring of another string.
Constraints:
1 <= words.length <= 1001 <= words[i].length <= 30words[i]contains only lowercase English letters.- All the strings of
wordsare unique.
TODO Approach
Write your approach here.
TODO Python
from collections import defaultdict as dd
class Solution:
def stringMatching(self, words: List[str]) -> List[str]:
words = reversed(sorted(words))
def squash(word, trie):
lst = []
for c in word:
lst.append([trie, ''])
for i in range(len(lst)):
lst[i][0] = lst[i][0][c]
lst[i][1] += c
if not lst[i][0]['ads']:
lst[i][0]['ads'] = {lst[i][1]: [word]}
else:
lst[i][0]['ads'][lst[i][1]].append(word)
def in_trie(word, trie):
for c in word:
if c not in trie:
return []
trie = trie[c]
return [(word, bw) for bw in trie['ads'][word]]
trie_maker = lambda: dd(trie_maker)
trie = trie_maker()
ans = []
for s in words:
ans += in_trie(s, trie)
squash(s, trie)
return ans
**
TODO C++
class Solution {
public:
vector<string> stringMatching(vector<string>& words) {
}
};