LeetCode #472 — HARD

Concatenated Words

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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 (not necessarily distinct) 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

Step 01

Brute Force Baseline

Problem summary: 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 (not necessarily distinct) in the given array.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Trie

Example 1

["cat","cats","catsdogcats","dog","dogcatsdog","hippopotamuses","rat","ratcatdogcat"]

Example 2

["cat","dog","catdog"]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #472: Concatenated Words
class Trie {
    Trie[] children = new Trie[26];
    boolean isEnd;

    void insert(String w) {
        Trie node = this;
        for (char c : w.toCharArray()) {
            c -= 'a';
            if (node.children[c] == null) {
                node.children[c] = new Trie();
            }
            node = node.children[c];
        }
        node.isEnd = true;
    }
}

class Solution {
    private Trie trie = new Trie();

    public List<String> findAllConcatenatedWordsInADict(String[] words) {
        Arrays.sort(words, (a, b) -> a.length() - b.length());
        List<String> ans = new ArrayList<>();
        for (String w : words) {
            if (dfs(w)) {
                ans.add(w);
            } else {
                trie.insert(w);
            }
        }
        return ans;
    }

    private boolean dfs(String w) {
        if ("".equals(w)) {
            return true;
        }
        Trie node = trie;
        for (int i = 0; i < w.length(); ++i) {
            int idx = w.charAt(i) - 'a';
            if (node.children[idx] == null) {
                return false;
            }
            node = node.children[idx];
            if (node.isEnd && dfs(w.substring(i + 1))) {
                return true;
            }
        }
        return false;
    }
}

// Accepted solution for LeetCode #472: Concatenated Words
type Trie struct {
	children [26]*Trie
	isEnd    bool
}

func newTrie() *Trie {
	return &Trie{}
}
func (this *Trie) insert(word string) {
	node := this
	for _, c := range word {
		c -= 'a'
		if node.children[c] == nil {
			node.children[c] = newTrie()
		}
		node = node.children[c]
	}
	node.isEnd = true
}

func findAllConcatenatedWordsInADict(words []string) (ans []string) {
	sort.Slice(words, func(i, j int) bool { return len(words[i]) < len(words[j]) })
	trie := newTrie()
	var dfs func(string) bool
	dfs = func(w string) bool {
		if w == "" {
			return true
		}
		node := trie
		for i, c := range w {
			c -= 'a'
			if node.children[c] == nil {
				return false
			}
			node = node.children[c]
			if node.isEnd && dfs(w[i+1:]) {
				return true
			}
		}
		return false
	}
	for _, w := range words {
		if dfs(w) {
			ans = append(ans, w)
		} else {
			trie.insert(w)
		}
	}
	return
}

# Accepted solution for LeetCode #472: Concatenated Words
class Trie:
    def __init__(self):
        self.children = [None] * 26
        self.is_end = False

    def insert(self, w):
        node = self
        for c in w:
            idx = ord(c) - ord('a')
            if node.children[idx] is None:
                node.children[idx] = Trie()
            node = node.children[idx]
        node.is_end = True


class Solution:
    def findAllConcatenatedWordsInADict(self, words: List[str]) -> List[str]:
        def dfs(w):
            if not w:
                return True
            node = trie
            for i, c in enumerate(w):
                idx = ord(c) - ord('a')
                if node.children[idx] is None:
                    return False
                node = node.children[idx]
                if node.is_end and dfs(w[i + 1 :]):
                    return True
            return False

        trie = Trie()
        ans = []
        words.sort(key=lambda x: len(x))
        for w in words:
            if dfs(w):
                ans.append(w)
            else:
                trie.insert(w)
        return ans

// Accepted solution for LeetCode #472: Concatenated Words
use std::collections::HashSet;
use std::iter::FromIterator;

impl Solution {
    pub fn find_all_concatenated_words_in_a_dict(words: Vec<String>) -> Vec<String> {
        let word_set: HashSet<String> = HashSet::from_iter(words.iter().cloned());
        let mut res = vec![];

        for w in words {
            if Self::dfs(&w, &word_set) {
                res.push(w);
            }
        }

        res
    }

    pub fn dfs(word: &str, word_set: &HashSet<String>) -> bool {
        for i in 1..word.len() {
            let (prefix, suffix) = (&word[..i], &word[i..]);
            if (word_set.contains(prefix) && word_set.contains(suffix))
                || (word_set.contains(prefix) && Self::dfs(suffix, word_set))
            {
                return true;
            }
        }
        false
    }
}

// Accepted solution for LeetCode #472: Concatenated Words
function findAllConcatenatedWordsInADict(words: string[]): string[] {
    let wordSet = new Set(words);
    let res: string[] = [];

    for (let w of words) {
        if (dfs(w)) {
            res.push(w);
        }
    }

    return res;

    function dfs(word: string): boolean {
        for (let i = 1; i < word.length; i++) {
            let prefix = word.slice(0, i);
            let suffix = word.slice(i);

            if (
                (wordSet.has(prefix) && wordSet.has(suffix)) ||
                (wordSet.has(prefix) && dfs(suffix))
            ) {
                return true;
            }
        }

        return false;
    }
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.