Substrings With Concatenation Of All Words Program Solution

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 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.
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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.

• Time: O(|\texttt{s}||\texttt{words}||\texttt{words[0]}|)
• Space: O(\Sigma |\texttt{words[i]}|)O(Σ∣words[i]∣)

SUBSTRING WITH CONCATENATION OF ALL WORDS Program Solution in C++

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 Program 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)
        ans.add(i);
    }

    return ans;
  }
}

SUBSTRING WITH CONCATENATION OF ALL WORDS Program 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

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