Longest Word in Dictionary

Longest Word in Dictionary - Problem

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.

Input & Output

Example 1 — Basic Building

$ Input: words = ["w","wo","wor","worl","world"]

› Output: "world"

💡 Note: "world" can be built one character at a time: "w" → "wo" → "wor" → "worl" → "world". All intermediate words exist in the dictionary.

Example 2 — Lexicographic Tie

$ Input: words = ["a","banana","app","appl","ap","apply","apple"]

› Output: "apple"

💡 Note: Both "apple" and "apply" can be built (a → ap → app → appl → apple/apply), but "apple" is lexicographically smaller.

Example 3 — No Valid Words

$ Input: words = ["hello","world"]

› Output: ""

💡 Note: Neither word can be built step by step. "hello" needs "h", "he", "hel", "hell" which don't exist. Same for "world".

Constraints

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

Visualization

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Asked in

G Google 42 a Amazon 28 M Microsoft 18 f Facebook 15

The key insight is that a word can only be built if all its prefixes exist in the dictionary. The optimal approach sorts words by length first, then processes them incrementally, ensuring each word only needs to check its immediate prefix. Time: O(n log n + n×m), Space: O(n×m).

Common Approaches

✓ One Pass Hash

⏱️ Time: N/A Space: N/A

Brute Force

⏱️ Time: O(n² × m) Space: O(n × m)

For each word, check if all its prefixes (from length 1 to length-1) exist in the dictionary. Keep track of the longest valid word with lexicographically smallest order.

DFS with Memoization

⏱️ Time: O(n × m²) Space: O(n × m)

For each word, use DFS to check if it can be built by checking all possible ways to split it into valid sub-words. Use memoization to cache results and avoid redundant computations.

Sort by Length First

⏱️ Time: O(n log n + n × m) Space: O(n × m)

Sort words by length first, then process from shortest to longest. A word is valid only if all its prefixes have been validated. This avoids redundant prefix checks.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

⚠ Quadratic Growth

Space Complexity

⚡ Linearithmic Space

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Medium Frequency

~25 min Avg. Time

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Smart Actions

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