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Link for the Problem – Substring with Concatenation of All Words– LeetCode Problem
Substring with Concatenation of All Words– LeetCode Problem
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 once, in 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]
Constraints:
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 { 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; } };
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; } }
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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