# Substring with Concatenation of All Words Leetcode Solution

In this post, we are going to solve the Substring with Concatenation of All Words Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

## Problem

You are given a string `s` and an array of strings `words`. All the strings of `words` are of the same length.

concatenated substring in `s` is a substring that 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 substring because it is not the concatenation of any permutation of `words`.

Return 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: Since words.length == 2 and words[i].length == 3, the concatenated substring has to be of length 6.
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.
The output order does not matter. Returning [9,0] is fine too.
```

Example 2:

```Input: s = "wordgoodgoodgoodbestword", words = ["word","good","best","word"]
Output: []
Explanation: Since words.length == 4 and words[i].length == 4, the concatenated substring has to be of length 16.
There is no substring of length 16 is s that is equal to the concatenation of any permutation of words.
We return an empty array.
```

Example 3:

```Input: s = "barfoofoobarthefoobarman", words = ["bar","foo","the"]
Output: [6,9,12]
Explanation: Since words.length == 3 and words[i].length == 3, the concatenated substring has to be of length 9.
The substring starting at 6 is "foobarthe". It is the concatenation of ["foo","bar","the"] which is a permutation of words.
The substring starting at 9 is "barthefoo". It is the concatenation of ["bar","the","foo"] which is a permutation of words.
The substring starting at 12 is "thefoobar". It is the concatenation of ["the","foo","bar"] which is a permutation of words.
```

Constraints:

• `1 <= s.length <= 104`
• `1 <= words.length <= 5000`
• `1 <= words[i].length <= 30`
• `s` and `words[i]` consist of lowercase English letters.

Now, lets see the leetcode solution of Substring with Concatenation of All Words Leetcode Solution.

### Substring with Concatenation of All Words Leetcode Solution in Python

```class Solution:
def findSubstring(self, s: str, words: List[str]) -> List[int]:
if len(s) == 0 or words == []:
return []

k = len(words)
n = len(words[0])
ans = []
count = Counter(words)

for i in range(len(s) - k * n + 1):
seen = defaultdict(int)
j = 0
while j < k:
word = s[i + j * n: i + j * n + n]
seen[word] += 1
if seen[word] > count[word]:
break
j += 1
if j == k:
ans.append(i)

return ans
```

### Substring with Concatenation of All WordsLeetcode Solutionin CPP

```class Solution {
public:
vector<int> findSubstring(string s, vector<string>& words) {
if (s.empty() || words.empty())
return {};

const int k = words.size();
const int n = words[0].length();
vector<int> ans;
unordered_map<string, int> count;

for (const string& word : words)
++count[word];

for (int i = 0; i < s.length() - k * n + 1; ++i) {
unordered_map<string, int> seen;
int j;
for (j = 0; j < k; ++j) {
const string& word = s.substr(i + j * n, n);
if (++seen[word] > count[word])
break;
}
if (j == k)
ans.push_back(i);
}

return ans;
}
};
```

### Substring with Concatenation of All Words Leetcode Solution in Java

```class Solution {
public List<Integer> findSubstring(String s, String[] words) {
if (s.isEmpty() || words.length == 0)
return new ArrayList<>();

final int k = words.length;
final int n = words[0].length();
List<Integer> ans = new ArrayList<>();
Map<String, Integer> count = new HashMap<>();

for (final String word : words)
count.put(word, count.getOrDefault(word, 0) + 1);

for (int i = 0; i <= s.length() - k * n; ++i) {
Map<String, Integer> seen = new HashMap<>();
int j = 0;
for (; j < k; ++j) {
final String word = s.substring(i + j * n, i + j * n + n);
seen.put(word, seen.getOrDefault(word, 0) + 1);
if (seen.get(word) > count.getOrDefault(word, 0))
break;
}
if (j == k)
}

return ans;
}
}
```

Note: This problem Substring with Concatenation of All Words is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.

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