Substring with Concatenation of All Words LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Substring with Concatenation of All Words in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the ProblemSubstring with Concatenation of All Words– LeetCode Problem

Substring with Concatenation of All Words– LeetCode Problem


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 oncein any order, and without any intervening characters.

You can return the answer in any order.

Example 1:

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:

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

Example 3:

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


  • 1 <= s.length <= 104
  • s consists of lower-case English letters.
  • 1 <= words.length <= 5000
  • 1 <= words[i].length <= 30
  • words[i] consists of lower-case English letters.
Substring with Concatenation of All Words– LeetCode Solutions
class Solution {
  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)

    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])
      if (j == k)

    return ans;
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))
      if (j == k)

    return ans;
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]:
        j += 1
      if j == k:

    return ans

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