Longest Word in Dictionary - Practice Coding Problems

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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Understanding the Visualization

Input

Array of dictionary words

Process

Find words that can be built character by character

Output

Longest buildable word (lexicographically smallest if tie)

Key Takeaway

🎯 Key Insight: A word can only be built if every single prefix exists in the dictionary

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

Approach	Time	Space	Notes
✓ DFS with Memoization	O(n × m²)	O(n × m)	Use DFS to check if a word can be built, with memoization to avoid recomputation
Brute Force	O(n² × m)	O(n × m)	Check each word to see if all its prefixes exist in the dictionary
Sort by Length First	O(n log n + n × m)	O(n × m)	Sort words by length, then build valid words incrementally

DFS with Memoization — Algorithm Steps

Create a word set for O(1) lookup
For each word, use DFS to check if it can be built
Use memoization to cache results of subproblems

Visualization

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Step-by-Step Walkthrough

Try Splits

DFS tries all ways to split the word

Cache Results

Memoization avoids recomputing subproblems

Find Longest

Return the longest buildable word

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MAX_WORDS 1000
#define MAX_LEN 31

char words[MAX_WORDS][MAX_LEN];
char memoKeys[MAX_WORDS * 10][MAX_LEN];
int memoValues[MAX_WORDS * 10];
int memoCount = 0;
int wordCount = 0;

int findWord(char* word) {
    for (int i = 0; i < wordCount; i++) {
        if (strcmp(words[i], word) == 0) {
            return 1;
        }
    }
    return 0;
}

int getMemo(char* word) {
    for (int i = 0; i < memoCount; i++) {
        if (strcmp(memoKeys[i], word) == 0) {
            return memoValues[i];
        }
    }
    return -1; // Not found
}

void setMemo(char* word, int value) {
    strcpy(memoKeys[memoCount], word);
    memoValues[memoCount] = value;
    memoCount++;
}

int canBuild(char* word) {
    int cached = getMemo(word);
    if (cached != -1) {
        return cached;
    }
    
    int len = strlen(word);
    if (len <= 1) {
        setMemo(word, 1);
        return 1;
    }
    
    // Try all possible splits
    for (int i = 1; i < len; i++) {
        char prefix[MAX_LEN], suffix[MAX_LEN];
        strncpy(prefix, word, i);
        prefix[i] = '\0';
        strcpy(suffix, word + i);
        
        if (findWord(prefix) && canBuild(suffix)) {
            setMemo(word, 1);
            return 1;
        }
    }
    
    setMemo(word, 0);
    return 0;
}

char* solution(char wordsArray[][MAX_LEN], int n) {
    static char result[MAX_LEN] = "";
    
    for (int i = 0; i < n; i++) {
        if (canBuild(wordsArray[i])) {
            int len = strlen(wordsArray[i]);
            if (len > strlen(result) || (len == strlen(result) && strcmp(wordsArray[i], result) < 0)) {
                strcpy(result, wordsArray[i]);
            }
        }
    }
    
    return result;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Simple parsing
    char* token = strtok(line, "[],\"\n ");
    while (token != NULL && wordCount < MAX_WORDS) {
        if (strlen(token) > 0) {
            strcpy(words[wordCount++], token);
        }
        token = strtok(NULL, "[],\"\n ");
    }
    
    char* result = solution(words, wordCount);
    printf("%s\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × m²)

For each word of length m, we check all possible splits with memoization avoiding recomputation

✓ Linear Growth

Space Complexity

O(n × m)

Word set storage plus memoization cache and recursion stack

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

DFS with Memoization — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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