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Home Code Solutions

Concatenated Words – LeetCode Solution

Concatenated Words - LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

BhautikBhalala by BhautikBhalala
September 10, 2022
Reading Time: 2 mins read
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Leetcode All Problems Solutions

Leetcode All Problems Solutions

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Table of Contents

  • Concatenated Words – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
  • C++ Concatenated Words LeetCode Solution
  • Java Concatenated Words LeetCode Solution
  • Python 3 Concatenated Words LeetCode Solution
    • Leave a comment below
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Concatenated Words – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

Given an array of strings words (without duplicates), return all the concatenated words in the given list of words.

A concatenated word is defined as a string that is comprised entirely of at least two shorter words in the given array.

 

Example 1:

Input: words = ["cat","cats","catsdogcats","dog","dogcatsdog","hippopotamuses","rat","ratcatdogcat"]
Output: ["catsdogcats","dogcatsdog","ratcatdogcat"]
Explanation: "catsdogcats" can be concatenated by "cats", "dog" and "cats"; 
"dogcatsdog" can be concatenated by "dog", "cats" and "dog"; 
"ratcatdogcat" can be concatenated by "rat", "cat", "dog" and "cat".

Example 2:

Input: words = ["cat","dog","catdog"]
Output: ["catdog"]

 

Constraints:

  • 1 <= words.length <= 104
  • 1 <= words[i].length <= 30
  • words[i] consists of only lowercase English letters.
  • All the strings of words are unique.
  • 1 <= sum(words[i].length) <= 105

C++ Concatenated Words LeetCode Solution


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vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        unordered_set<string> s(words.begin(), words.end());
        vector<string> res;
        for (auto w : words) {
            int n = w.size();
            vector<int> dp(n+1);
            dp[0] = 1;
            for (int i = 0; i < n; i++) {
                if (dp[i] == 0) continue;
                for (int j = i+1; j <= n; j++) {
                    if (j - i < n && s.count(w.substr(i, j-i))) dp[j] = 1;
                }
                if (dp[n]) { res.push_back(w); break; }
            }
        }
        return res;
    }

Java Concatenated Words LeetCode Solution


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public class Solution {
    public static List<String> findAllConcatenatedWordsInADict(String[] words) {
        List<String> result = new ArrayList<>();
        Set<String> preWords = new HashSet<>();
        Arrays.sort(words, new Comparator<String>() {
            public int compare (String s1, String s2) {
                return s1.length() - s2.length();
            }
        });
        
        for (int i = 0; i < words.length; i++) {
            if (canForm(words[i], preWords)) {
                result.add(words[i]);
            }
            preWords.add(words[i]);
        }
        
        return result;
    }
	
    private static boolean canForm(String word, Set<String> dict) {
        if (dict.isEmpty()) return false;
	boolean[] dp = new boolean[word.length() + 1];
	dp[0] = true;
	for (int i = 1; i <= word.length(); i++) {
	    for (int j = 0; j < i; j++) {
		if (!dp[j]) continue;
		if (dict.contains(word.substring(j, i))) {
		    dp[i] = true;
		    break;
		}
	    }
	}
	return dp[word.length()];
    }
}

 



Python 3 Concatenated Words LeetCode Solution


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class Solution(object):
    def findAllConcatenatedWordsInADict(self, words):
        """
        :type words: List[str]
        :rtype: List[str]
        """
        d = set(words)
        
        def dfs(word):
            for i in range(1, len(word)):
                prefix = word[:i]
                suffix = word[i:]
                
                if prefix in d and suffix in d:
                    return True
                if prefix in d and dfs(suffix):
                    return True
                if suffix in d and dfs(prefix):
                    return True
            
            return False
        
        res = []
        for word in words:
            if dfs(word):
                res.append(word)
        
        return res



 

Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264


Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195


Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140


Stack-133
Heap (Priority Queue)-117
Design-116
Graph-108
Simulation-103
Prefix Sum-96


Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
Union Find-66


Ordered Set-48
Monotonic Stack-47
Recursion-43
Trie-41
Binary Search Tree-40
Divide and Conquer-40


Enumeration-39
Bitmask-37
Queue-33
Memoization-32
Topological Sort-31


Geometry-30
Segment Tree-27
Game Theory-24
Hash Function-24
Binary Indexed Tree-21


Interactive-18
Data Stream-17
String Matching-17
Rolling Hash-17
Shortest Path-16


Number Theory-16
Combinatorics-15
Randomized-12
Monotonic Queue-9
Iterator-9
Merge Sort-9


Concurrency-9
Doubly-Linked List-8
Brainteaser-8
Probability and Statistics-7
Quickselect-7


Bucket Sort-6
Suffix Array-6
Minimum Spanning Tree-5
Counting Sort-5
Shell-4


Line Sweep-4
Reservoir Sampling-4
Eulerian Circuit-3
Radix Sort-3


Strongly Connected Componen-t2
Rejection Sampling-2
Biconnected Component-1

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Tags: Cc++14Concatenated Wordsfull solutionGojavajava 15java 7java 8java8javascriptkotlinLeetCodeLeetCodeSolutionpypy 3Python 2python 3rubyrustSolution
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