30. Substring with Concatenation of All Words
Description
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.
Solutions
Solution 1: Hash Table + Sliding Window
We use a hash table $cnt$ to count the number of times each word appears in $words$, and use a hash table $cnt1$ to count the number of times each word appears in the current sliding window. We denote the length of the string $s$ as $m$, the number of words in the string array $words$ as $n$, and the length of each word as $k$.
We can enumerate the starting point $i$ of the sliding window, where $0 \lt i < k$. For each starting point, we maintain a sliding window with the left boundary as $l$, the right boundary as $r$, and the number of words in the sliding window as $t$. Additionally, we use a hash table $cnt1$ to count the number of times each word appears in the sliding window.
Each time, we extract the string $s[r:r+k]$. If $s[r:r+k]$ is not in the hash table $cnt$, it means that the words in the current sliding window are not valid. We update the left boundary $l$ to $r$, clear the hash table $cnt1$, and reset the word count $t$ to 0. If $s[r:r+k]$ is in the hash table $cnt$, it means that the words in the current sliding window are valid. We increase the word count $t$ by 1, and increase the count of $s[r:r+k]$ in the hash table $cnt1$ by 1. If $cnt1[s[r:r+k]]$ is greater than $cnt[s[r:r+k]]$, it means that $s[r:r+k]$ appears too many times in the current sliding window. We need to move the left boundary $l$ to the right until $cnt1[s[r:r+k]] = cnt[s[r:r+k]]$. If $t = n$, it means that the words in the current sliding window are exactly valid, and we add the left boundary $l$ to the answer array.
The time complexity is $O(m \times k)$, and the space complexity is $O(n \times k)$. Here, $m$ and $n$ are the lengths of the string $s$ and the string array $words$ respectively, and $k$ is the length of the words in the string array $words$.
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Solution 2
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