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‎ConcatenatedWords/concatenated_words.dart

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/*
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-* 472. Concatenated Words *-
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Given an array of strings words (without duplicates), return all the concatenated words in the given list of words.
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A concatenated word is defined as a string that is comprised entirely of at least two shorter words in the given array.
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Example 1:
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Input: words = ["cat","cats","catsdogcats","dog","dogcatsdog","hippopotamuses","rat","ratcatdogcat"]
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Output: ["catsdogcats","dogcatsdog","ratcatdogcat"]
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Explanation: "catsdogcats" can be concatenated by "cats", "dog" and "cats";
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"dogcatsdog" can be concatenated by "dog", "cats" and "dog";
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"ratcatdogcat" can be concatenated by "rat", "cat", "dog" and "cat".
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Example 2:
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Input: words = ["cat","dog","catdog"]
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Output: ["catdog"]
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Constraints:
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1 <= words.length <= 104
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1 <= words[i].length <= 30
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words[i] consists of only lowercase English letters.
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All the strings of words are unique.
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1 <= sum(words[i].length) <= 105
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*/
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import 'dart:collection';
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class A {
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bool checkConcatenate(String word, Set<String> s) {
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for (int i = 1; i < word.length; i++) {
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String prefixWord = word.substring(0, i);
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String suffixWord = word.substring(i, word.length);
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if (s.contains(prefixWord) &&
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(s.contains(suffixWord) || checkConcatenate(suffixWord, s)))
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return true;
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}
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return false;
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}
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List<String> findAllConcatenatedWordsInADict(List<String> words) {
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HashSet<String> s = HashSet();
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List<String> concatenateWords = [];
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for (String word in words) s.add(word);
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for (String word in words) {
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if (checkConcatenate(word, s) == true) concatenateWords.add(word);
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}
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return concatenateWords;
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}
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}
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class B {
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List<String> findAllConcatenatedWordsInADict(List<String> words) {
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HashSet<String> wordsSet = HashSet();
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for (String word in words) wordsSet.add(word);
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List<String> res = [];
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for (String word in words) {
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int n = word.length;
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List<int> dp = List.filled(n + 1, 0);
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dp[0] = 1;
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for (int i = 0; i < n; i++) {
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if (dp[i] != 0) continue;
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for (int j = i + 1; j <= n; j++) {
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if (j - i < n && wordsSet.contains(word.substring(i, j - i))) {
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dp[j] = 1;
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}
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}
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if (dp[n] == 0) {
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res.add(word);
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break;
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}
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}
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}
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return res;
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}
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}
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class C {
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List<String> findAllConcatenatedWordsInADict(List<String> words) {
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List<String> result = [];
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HashSet<String> set = HashSet();
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for (String word in words) set.add(word);
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for (String word in words) helper(result, set, word);
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return result;
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}
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void helper(List<String> list, HashSet<String> set, String word) {
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if (word.length == 0) return;
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List<bool> dp = List.filled(word.length + 1, false);
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dp[0] = true;
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for (int i = 0; i < word.length; i++) {
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if (!dp[i]) continue;
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for (int j = i + 1; j < dp.length; j++) {
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if (i == 0 && j == word.length) continue;
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if (set.contains(word.substring(i, j))) dp[j] = true;
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}
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}
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if (dp[dp.length - 1]) list.add(word);
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}
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}

‎ConcatenatedWords/concatenated_words.go

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package main
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type Trie struct {
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Children map[string]*Trie // important to make it a pointer to modify the Children.Trie value
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End bool
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}
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func NewTrie() *Trie {
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return &Trie{
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Children: make(map[string]*Trie),
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End: false,
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}
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}
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func addWord(t *Trie, word string) {
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var current *Trie = t
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for i := 0; i < len(word); i++ {
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letter := string(word[i])
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if _, ok := current.Children[letter]; !ok {
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current.Children[letter] = NewTrie()
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}
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current = current.Children[letter]
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}
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current.End = true
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}
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func searchConcat(root *Trie, word string, index int, count int) bool {
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// base case
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if index >= len(word) {
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return count >= 2
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}
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var current *Trie = root
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for i := index; i < len(word); i++ {
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letter := string(word[i])
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if _, ok := current.Children[letter]; !ok {
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break
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}
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current = current.Children[letter]
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// when we reach the end, 2 choices
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// restart or continue
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if current.End {
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// restart
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atRestart := searchConcat(root, word, i+1, count+1)
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if atRestart {
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return true
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} else {
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// continue down the branch
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continue
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}
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}
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}
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return false
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}
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func findAllConcatenatedWordsInADict(words []string) []string {
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var trie *Trie = NewTrie()
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for _, word := range words {
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addWord(trie, word)
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}
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var validWords []string = []string{}
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for _, word := range words {
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isFound := searchConcat(trie, word, 0, 0)
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if isFound {
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validWords = append(validWords, word)
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}
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}
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return validWords
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}

‎ConcatenatedWords/concatenated_words.md

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# 🔥 2 Approaches 🔥 || Simple Fast and Easy || with Explanation
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## Solution - Recursion - Depth First Search
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### Approach for this Problem
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Create an empty set 's' to store all the words in the given array of strings.
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Iterate through the array of strings and insert each word into the set 's'.
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Create an empty vector 'concatenateWords' to store all the concatenated words.
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Iterate through the array of strings again, for each word, check if it is a concatenated word using the function 'checkConcatenate(word)'.
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In the 'checkConcatenate(word)' function, use a for loop to iterate through each substring of the word, starting from index 1 to the second last index of the word.
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For each substring, check if the prefix and suffix of the substring exists in the set 's'.
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If the prefix and suffix both exist in the set 's', then return true, indicating that the word is a concatenated word.
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If the function 'checkConcatenate(word)' returns true, then insert the word into the 'concatenateWords' vector.
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Return the 'concatenateWords' vector.
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### Time Complexity and Space Complexity
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- Time complexity: O(n^2*m) //where n is the number of words in the input array and m is the average length of the words.
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- Space complexity: O(n*m) //where n is the number of words in the input array and m is the average length of the words.
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### Code
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```dart
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import 'dart:collection';
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class Solution {
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bool checkConcatenate(String word, Set<String> s) {
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for (int i = 1; i < word.length; i++) {
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String prefixWord = word.substring(0, i);
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String suffixWord = word.substring(i, word.length);
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if (s.contains(prefixWord) &&
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(s.contains(suffixWord) || checkConcatenate(suffixWord, s)))
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return true;
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}
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return false;
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}
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List<String> findAllConcatenatedWordsInADict(List<String> words) {
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HashSet<String> s = HashSet();
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List<String> concatenateWords = [];
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for (String word in words) s.add(word);
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for (String word in words) {
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if (checkConcatenate(word, s) == true) concatenateWords.add(word);
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}
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return concatenateWords;
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}
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}
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```
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## Solution - 2 DP
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### Code
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```dart
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class Solution {
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List<String> findAllConcatenatedWordsInADict(List<String> words) {
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List<String> result = [];
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HashSet<String> set = HashSet();
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for (String word in words) set.add(word);
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for (String word in words) helper(result, set, word);
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return result;
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}
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void helper(List<String> list, HashSet<String> set, String word) {
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if (word.length == 0) return;
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List<bool> dp = List.filled(word.length + 1, false);
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dp[0] = true;
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for (int i = 0; i < word.length; i++) {
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if (!dp[i]) continue;
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for (int j = i + 1; j < dp.length; j++) {
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if (i == 0 && j == word.length) continue;
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if (set.contains(word.substring(i, j))) dp[j] = true;
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}
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}
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if (dp[dp.length - 1]) list.add(word);
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}
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}
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```

‎README.md

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- [**909.** Snakes and Ladders](SnakesAndLadders/snakes_and_ladders.dart)
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- [**2359.** Find Closest Node to Given Two Nodes](FindClosestNodeToGivenTwoNodes/find_closest_node_to_given_two_nodes.dart)
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- [**787.** Cheapest Flights Within K Stops](CheapestFlightsWithinKStops/cheapest_flights_within_k_stops.dart)
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- [**472.** Concatenated Words](ConcatenatedWords/concatenated_words.dart)
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## Reach me via
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