LeetCode #720 — MEDIUM

Longest Word in Dictionary

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

Given an array of strings words representing an English Dictionary, return the longest word in words that can be built one character at a time by other words in words.

If there is more than one possible answer, return the longest word with the smallest lexicographical order. If there is no answer, return the empty string.

Note that the word should be built from left to right with each additional character being added to the end of a previous word.

Example 1:

Input: words = ["w","wo","wor","worl","world"]
Output: "world"
Explanation: The word "world" can be built one character at a time by "w", "wo", "wor", and "worl".

Example 2:

Input: words = ["a","banana","app","appl","ap","apply","apple"]
Output: "apple"
Explanation: Both "apply" and "apple" can be built from other words in the dictionary. However, "apple" is lexicographically smaller than "apply".

Constraints:

1 <= words.length <= 1000
1 <= words[i].length <= 30
words[i] consists of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: Given an array of strings words representing an English Dictionary, return the longest word in words that can be built one character at a time by other words in words. If there is more than one possible answer, return the longest word with the smallest lexicographical order. If there is no answer, return the empty string. Note that the word should be built from left to right with each additional character being added to the end of a previous word.

Baseline thinking

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

Pattern signal: Array · Hash Map · Trie

Example 1

["w","wo","wor","worl","world"]

Example 2

["a","banana","app","appl","ap","apply","apple"]

Core Insight

What unlocks the optimal approach

For every word in the input list, we can check whether all prefixes of that word are in the input list by using a Set.

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

Upper-end input sizes

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 #720: Longest Word in Dictionary
class Trie {
    private Trie[] children = new Trie[26];
    private boolean isEnd = false;

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

    public boolean search(String w) {
        Trie node = this;
        for (char c : w.toCharArray()) {
            int idx = c - 'a';
            if (node.children[idx] == null || !node.children[idx].isEnd) {
                return false;
            }
            node = node.children[idx];
        }
        return true;
    }
}

class Solution {
    public String longestWord(String[] words) {
        Trie trie = new Trie();
        for (String w : words) {
            trie.insert(w);
        }
        String ans = "";
        for (String w : words) {
            if (trie.search(w)
                && (ans.length() < w.length()
                    || (ans.length() == w.length() && w.compareTo(ans) < 0))) {
                ans = w;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #720: Longest Word in Dictionary
type Trie struct {
	children [26]*Trie
	isEnd    bool
}

func (t *Trie) insert(w string) {
	node := t
	for i := 0; i < len(w); i++ {
		idx := w[i] - 'a'
		if node.children[idx] == nil {
			node.children[idx] = &Trie{}
		}
		node = node.children[idx]
	}
	node.isEnd = true
}

func (t *Trie) search(w string) bool {
	node := t
	for i := 0; i < len(w); i++ {
		idx := w[i] - 'a'
		if node.children[idx] == nil || !node.children[idx].isEnd {
			return false
		}
		node = node.children[idx]
	}
	return true
}

func longestWord(words []string) string {
	trie := &Trie{}
	for _, w := range words {
		trie.insert(w)
	}

	ans := ""
	for _, w := range words {
		if trie.search(w) && (len(ans) < len(w) || (len(ans) == len(w) && w < ans)) {
			ans = w
		}
	}
	return ans
}

# Accepted solution for LeetCode #720: Longest Word in Dictionary
class Trie:
    def __init__(self):
        self.children: List[Optional[Trie]] = [None] * 26
        self.is_end = False

    def insert(self, w: str):
        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

    def search(self, w: str) -> bool:
        node = self
        for c in w:
            idx = ord(c) - ord("a")
            if node.children[idx] is None:
                return False
            node = node.children[idx]
            if not node.is_end:
                return False
        return True


class Solution:
    def longestWord(self, words: List[str]) -> str:
        trie = Trie()
        for w in words:
            trie.insert(w)
        ans = ""
        for w in words:
            if trie.search(w) and (
                len(ans) < len(w) or (len(ans) == len(w) and ans > w)
            ):
                ans = w
        return ans

// Accepted solution for LeetCode #720: Longest Word in Dictionary
struct Trie {
    children: [Option<Box<Trie>>; 26],
    is_end: bool,
}

impl Trie {
    fn new() -> Self {
        Trie {
            children: Default::default(),
            is_end: false,
        }
    }

    fn insert(&mut self, w: &str) {
        let mut node = self;
        for c in w.chars() {
            let idx = (c as usize) - ('a' as usize);
            if node.children[idx].is_none() {
                node.children[idx] = Some(Box::new(Trie::new()));
            }
            node = node.children[idx].as_mut().unwrap();
        }
        node.is_end = true;
    }

    fn search(&self, w: &str) -> bool {
        let mut node = self;
        for c in w.chars() {
            let idx = (c as usize) - ('a' as usize);
            if node.children[idx].is_none() || !node.children[idx].as_ref().unwrap().is_end {
                return false;
            }
            node = node.children[idx].as_ref().unwrap();
        }
        true
    }
}

impl Solution {
    pub fn longest_word(words: Vec<String>) -> String {
        let mut trie = Trie::new();
        for w in &words {
            trie.insert(w);
        }

        let mut ans = String::new();
        for w in words {
            if trie.search(&w) && (ans.len() < w.len() || (ans.len() == w.len() && w < ans)) {
                ans = w;
            }
        }
        ans
    }
}

// Accepted solution for LeetCode #720: Longest Word in Dictionary
class Trie {
    children: (Trie | null)[] = new Array(26).fill(null);
    isEnd: boolean = false;

    insert(w: string): void {
        let node: Trie = this;
        for (let i = 0; i < w.length; i++) {
            const idx: number = w.charCodeAt(i) - 'a'.charCodeAt(0);
            if (node.children[idx] === null) {
                node.children[idx] = new Trie();
            }
            node = node.children[idx]!;
        }
        node.isEnd = true;
    }

    search(w: string): boolean {
        let node: Trie = this;
        for (let i = 0; i < w.length; i++) {
            const idx: number = w.charCodeAt(i) - 'a'.charCodeAt(0);
            if (node.children[idx] === null || !node.children[idx]!.isEnd) {
                return false;
            }
            node = node.children[idx]!;
        }
        return true;
    }
}

function longestWord(words: string[]): string {
    const trie = new Trie();
    for (const w of words) {
        trie.insert(w);
    }

    let ans = '';
    for (const w of words) {
        if (trie.search(w) && (ans.length < w.length || (ans.length === w.length && w < ans))) {
            ans = w;
        }
    }
    return ans;
}

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(L)

Space

O(L)

Approach Breakdown

HASH SET

O(N × L) time

O(N × L) space

Store all N words in a hash set. Each insert/lookup hashes the entire word of length L, giving O(L) per operation. Prefix queries require checking every stored word against the prefix — O(N × L) per prefix search. Space is O(N × L) for storing all characters.

TRIE

O(L) time

O(N × L) space

Each operation (insert, search, prefix) takes O(L) time where L is the word length — one node visited per character. Total space is bounded by the sum of all stored word lengths. Tries win over hash sets when you need prefix matching: O(L) prefix search vs. checking every stored word.

Shortcut: One node per character → O(L) per operation. Prefix queries are what make tries worthwhile.

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.