String Matching in an Array

String Matching in an Array - Problem

Given an array of string words, return all strings in words that are a substring of another word. You can return the answer in any order.

A substring is a contiguous sequence of characters within a string.

Input & Output

Example 1 — Basic Case

$ Input: words = ["mass","as","hero","superhero"]

› Output: ["as","hero"]

💡 Note: "as" is a substring of "mass" and "hero" is a substring of "superhero". Return ["as","hero"].

Example 2 — No Substrings

$ Input: words = ["leetcode","et","code"]

› Output: ["et","code"]

💡 Note: "et" is a substring of "leetcode" and "code" is also a substring of "leetcode".

Example 3 — No Matches

$ Input: words = ["blue","green","red"]

› Output: []

💡 Note: None of the words are substrings of another word, so return empty array.

Constraints

1 ≤ words.length ≤ 1000
1 ≤ words[i].length ≤ 30
words[i] contains only lowercase English letters.

Visualization

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Asked in

G Google 25 f Facebook 20 a Amazon 15

The key insight is that we need to compare every word with every other word to find substring relationships. The brute force approach works well here since we must check all pairs anyway. Best approach uses nested loops: Time: O(n²×m), Space: O(k).

Common Approaches

✓ Brute Force - Check All Pairs

⏱️ Time: O(n²×m) Space: O(k)

Compare every word with every other word to see if one contains the other as a substring. Use the built-in string contains method for substring checking.

Optimized - Sort by Length First

⏱️ Time: O(n²×m) Space: O(k)

Since a substring must be shorter than or equal to its parent string, we can sort by length first and only check shorter words against longer ones. This reduces some unnecessary comparisons.

Brute Force - Check All Pairs — Algorithm Steps

Iterate through each word in the array
For each word, check against all other words
If current word is found as substring in another word, add to result

Visualization

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Step-by-Step Walkthrough

Compare Each Word

Check if current word exists in any other word

Substring Check

Use built-in contains/find method

Collect Results

Add matching substrings to result array

Code -

solution.c — C

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

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    // Simple parsing for JSON array of strings
    char words[100][100];
    int wordCount = 0;
    char* ptr = line + 1; // Skip '['
    
    while (*ptr && *ptr != ']') {
        while (*ptr == ' ' || *ptr == ',' || *ptr == '"') ptr++;
        if (*ptr == ']') break;
        
        int i = 0;
        while (*ptr && *ptr != '"' && *ptr != ']') {
            words[wordCount][i++] = *ptr++;
        }
        words[wordCount][i] = 0;
        if (strlen(words[wordCount]) > 0) wordCount++;
        
        while (*ptr && *ptr != ',' && *ptr != ']') ptr++;
    }
    
    // Find substrings
    char result[100][100];
    int resultCount = 0;
    
    for (int i = 0; i < wordCount; i++) {
        for (int j = 0; j < wordCount; j++) {
            if (i != j && strstr(words[j], words[i]) != NULL) {
                strcpy(result[resultCount++], words[i]);
                break;
            }
        }
    }
    
    // Output as JSON array
    printf("[");
    for (int i = 0; i < resultCount; i++) {
        if (i > 0) printf(",");
        printf("\"%s\"", result[i]);
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²×m)

Nested loops over n words, each substring check takes O(m) time

⚠ Quadratic Growth

Space Complexity

O(k)

Only store result list with k matching substrings

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

856 Likes

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check All Pairs — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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