0030: Substring with Concatenation of All Words
Problem Statement
You are given a string s
and an array of strings words
. All the strings of words
are of the same length.
A concatenated string is a string that exactly contains all the strings of any permutation of words
concatenated.
- For example, if
words = ["ab","cd","ef"]
, then"abcdef"
,"abefcd"
,"cdabef"
,"cdefab"
,"efabcd"
, and"efcdab"
are all concatenated strings."acdbef"
is not a concatenated string because it is not the concatenation of any permutation ofwords
.
Return an array of the starting indices of all the concatenated substrings in s
. You can return the answer in any order.
Example 1:
Input: s = "barfoothefoobarman", words = ["foo","bar"]
Output: [0,9]
Explanation:
The substring starting at 0 is "barfoo"
. It is the concatenation of ["bar","foo"]
which is a permutation of words
.
The substring starting at 9 is "foobar"
. It is the concatenation of ["foo","bar"]
which is a permutation of words
.
Example 2:
Input: s = "wordgoodgoodgoodbestword", words = ["word","good","best","word"]
Output: []
Explanation:
There is no concatenated substring.
Example 3:
Input: s = "barfoofoobarthefoobarman", words = ["bar","foo","the"]
Output: [6,9,12]
Explanation:
The substring starting at 6 is "foobarthe"
. It is the concatenation of ["foo","bar","the"]
.
The substring starting at 9 is "barthefoo"
. It is the concatenation of ["bar","the","foo"]
.
The substring starting at 12 is "thefoobar"
. It is the concatenation of ["the","foo","bar"]
.
Constraints:
1 <= s.length <= 104
1 <= words.length <= 5000
1 <= words[i].length <= 30
s
andwords[i]
consist of lowercase English letters.
Code Solution
from collections import Counter
class Solution:
def findSubstring(self, s: str, words: List[str]) -> List[int]:
N = len(s)
wn = len(words)
ws = len(words[0])
out = []
tgt = dict(Counter(words))
for offset in range(ws):
for ix in range(offset, N - (ws * wn) + 1, ws):
src = [s[ix + jx:ix + jx + ws] for jx in range(0, (ws * wn), ws)]
src = dict(Counter(src))
if src == tgt:
out.append(ix)
return out