# Substring with Concatenation of All Words Problem

## Description

LeetCode Problem 30.

You are given a string s and an array of strings words of the same length. Return all starting indices of substring(s) in s that is a concatenation of each word in words exactly once, in any order, and without any intervening characters.

You can return the answer in any order.

Example 1:

``````1
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Input: s = "barfoothefoobarman", words = ["foo","bar"]
Output: [0,9]
Explanation: Substrings starting at index 0 and 9 are "barfoo" and "foobar" respectively.
The output order does not matter, returning [9,0] is fine too.
``````

Example 2:

``````1
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Input: s = "wordgoodgoodgoodbestword", words = ["word","good","best","word"]
Output: []
``````

Example 3:

``````1
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Input: s = "barfoofoobarthefoobarman", words = ["bar","foo","the"]
Output: [6,9,12]
``````

Constraints:

• 1 <= s.length <= 10^4
• s consists of lower-case English letters.
• 1 <= words.length <= 5000
• 1 <= words[i].length <= 30
• words[i] consists of lower-case English letters.

## Sample C++ Code

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class Solution {
public:
vector findSubstring(string s, vector& words) {
unordered_map<string, int> counts;
for (string word : words)
counts[word]++;
int n = s.length(), num = words.size(), len = words[0].length();
vector indexes;
for (int i = 0; i < n - num * len + 1; i++) {
unordered_map<string, int> seen;
int j = 0;
for (; j < num; j++) {
string word = s.substr(i + j * len, len);
if (counts.find(word) != counts.end()) {
seen[word]++;
if (seen[word] > counts[word])
break;
}
else break;
}
if (j == num) indexes.push_back(i);
}
return indexes;
}
};
``````