Before and After Puzzle

Before and After Puzzle - Problem

Given a list of phrases, generate a list of Before and After puzzles.

A phrase is a string that consists of lowercase English letters and spaces only. No space appears in the start or the end of a phrase. There are no consecutive spaces in a phrase.

Before and After puzzles are phrases that are formed by merging two phrases where the last word of the first phrase is the same as the first word of the second phrase. Note that only the last word of the first phrase and the first word of the second phrase are merged in this process.

Return the Before and After puzzles that can be formed by every two phrases phrases[i] and phrases[j] where i != j. Note that the order of matching two phrases matters, we want to consider both orders.

You should return a list of distinct strings sorted lexicographically, after removing all duplicate phrases in the generated Before and After puzzles.

Input & Output

Example 1 — Basic Connection

$ Input: phrases = ["writing code", "code review"]

› Output: ["writing code review"]

💡 Note: The last word "code" of first phrase matches the first word "code" of second phrase. Merge by removing the duplicate: "writing code" + "review" = "writing code review"

Example 2 — Multiple Connections

$ Input: phrases = ["mission statement", "a quick", "quick brown", "brown fox", "fox jumps"]

› Output: ["a quick brown fox", "a quick brown fox jumps", "quick brown fox jumps"]

💡 Note: Multiple chains possible: "a quick" → "quick brown" → "brown fox" → "fox jumps" creates several valid before-and-after puzzles

Example 3 — No Connections

$ Input: phrases = ["hello world", "good morning", "nice day"]

› Output: []

💡 Note: No phrase's last word matches any other phrase's first word, so no before-and-after puzzles can be formed

Constraints

1 ≤ phrases.length ≤ 100
1 ≤ phrases[i].length ≤ 100
phrases[i] consists of lowercase English letters and spaces
No phrase has leading or trailing spaces
No phrase has consecutive spaces

Visualization

Tap to expand

Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to recognize that we need to match last words with first words efficiently. The best approach uses hash maps to group phrases by their first and last words, enabling O(1) lookups instead of O(n²) comparisons. Time: O(n×m + k×m), Space: O(n×m) where n is phrases count, m is average phrase length, and k is result count.

Common Approaches

✓ Frequency Count

⏱️ Time: N/A Space: N/A

Brute Force

⏱️ Time: O(n² × m) Space: O(k × m)

For each pair of phrases, extract their first and last words. If the last word of the first phrase matches the first word of the second phrase, merge them by combining all words except the duplicate connecting word.

Hash Map Optimization

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

Instead of checking every pair, group phrases by their first and last words using hash maps. For each phrase with last word W, quickly find all phrases starting with W using the hash map lookup.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

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

#define MAX_PHRASES 100
#define MAX_LENGTH 1000
#define MAX_RESULTS 10000

static char phrases[MAX_PHRASES][MAX_LENGTH];
static char results[MAX_RESULTS][MAX_LENGTH];
static int result_count = 0;

// Extract first word from phrase
void getFirstWord(const char* phrase, char* first) {
    int i = 0;
    while (phrase[i] && phrase[i] != ' ') {
        first[i] = phrase[i];
        i++;
    }
    first[i] = '\0';
}

// Extract last word from phrase
void getLastWord(const char* phrase, char* last) {
    int len = strlen(phrase);
    int start = len - 1;
    
    // Find start of last word
    while (start >= 0 && phrase[start] != ' ') {
        start--;
    }
    start++; // Move to first character of last word
    
    strcpy(last, phrase + start);
}

// Get all words except the last one (concatenated without spaces)
void getPrefix(const char* phrase, char* prefix) {
    int len = strlen(phrase);
    int end = len - 1;
    
    // Find end of prefix (before last word)
    while (end >= 0 && phrase[end] != ' ') {
        end--;
    }
    
    prefix[0] = '\0';
    if (end >= 0) {
        // Extract words and concatenate without spaces
        int i = 0, j = 0;
        while (i < end) {
            if (phrase[i] != ' ') {
                prefix[j++] = phrase[i];
            }
            i++;
        }
        prefix[j] = '\0';
    }
}

// Get all words except the first one (concatenated without spaces)
void getSuffix(const char* phrase, char* suffix) {
    int i = 0;
    
    // Skip first word
    while (phrase[i] && phrase[i] != ' ') {
        i++;
    }
    
    suffix[0] = '\0';
    if (phrase[i] == ' ') {
        i++; // Skip the space after first word
        int j = 0;
        while (phrase[i]) {
            if (phrase[i] != ' ') {
                suffix[j++] = phrase[i];
            }
            i++;
        }
        suffix[j] = '\0';
    }
}

// Check if result already exists
int exists(const char* result) {
    for (int i = 0; i < result_count; i++) {
        if (strcmp(results[i], result) == 0) {
            return 1;
        }
    }
    return 0;
}

// Compare function for qsort
int compare(const void* a, const void* b) {
    return strcmp((char*)a, (char*)b);
}

int main() {
    char input[MAX_LENGTH];
    fgets(input, sizeof(input), stdin);
    
    // Parse JSON array of phrases
    int phrase_count = 0;
    char* ptr = input;
    
    // Skip opening bracket and whitespace
    while (*ptr && (*ptr == '[' || *ptr == ' ' || *ptr == '\n')) ptr++;
    
    while (*ptr && *ptr != ']') {
        if (*ptr == '"') {
            ptr++; // Skip opening quote
            int i = 0;
            while (*ptr && *ptr != '"') {
                phrases[phrase_count][i++] = *ptr++;
            }
            phrases[phrase_count][i] = '\0';
            phrase_count++;
            
            if (*ptr == '"') ptr++; // Skip closing quote
        }
        
        // Skip comma and whitespace
        while (*ptr && (*ptr == ',' || *ptr == ' ')) ptr++;
    }
    
    // Find all before and after puzzles
    for (int i = 0; i < phrase_count; i++) {
        for (int j = 0; j < phrase_count; j++) {
            if (i == j) continue;
            
            char last_word[MAX_LENGTH], first_word[MAX_LENGTH];
            getLastWord(phrases[i], last_word);
            getFirstWord(phrases[j], first_word);
            
            if (strcmp(last_word, first_word) == 0) {
                char prefix[MAX_LENGTH], suffix[MAX_LENGTH];
                char merged[MAX_LENGTH];
                
                getPrefix(phrases[i], prefix);
                getSuffix(phrases[j], suffix);
                
                // Build merged phrase (concatenate without spaces)
                merged[0] = '\0';
                if (strlen(prefix) > 0) {
                    strcat(merged, prefix);
                }
                strcat(merged, last_word);
                if (strlen(suffix) > 0) {
                    strcat(merged, suffix);
                }
                
                if (!exists(merged)) {
                    strcpy(results[result_count], merged);
                    result_count++;
                }
            }
        }
    }
    
    // Sort results
    qsort(results, result_count, sizeof(results[0]), compare);
    
    // Output results
    printf("[");
    for (int i = 0; i < result_count; i++) {
        if (i > 0) printf(",");
        printf("\"%s\"", results[i]);
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

⚡ Linearithmic Space

18.5K Views

Medium Frequency

~25 min Avg. Time

487 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen