Introduction

In this blog post, we will explore how to solve the LeetCode problem "1408" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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. Example 1: Input: words = ["mass","as","hero","superhero"] Output: ["as","hero"] Explanation: "as" is substring of "mass" and "hero" is substring of "superhero". ["hero","as"] is also a valid answer. Example 2: Input: words = ["leetcode","et","code"] Output: ["et","code"] Explanation: "et", "code" are substring of "leetcode". Example 3: Input: words = ["blue","green","bu"] Output: [] Explanation: No string of words is substring of another string. Constraints: 1 <= words.length <= 100 1 <= words[i].length <= 30 words[i] contains only lowercase English letters. All the strings of words are unique.

Explanation

Here's the approach to solve this problem:

• Iterate and Compare: The core idea is to iterate through each word in the input array and check if it is a substring of any other word in the array.
• Avoid Self-Comparison: Make sure to not compare a word against itself.

• Efficient Substring Check: Use Python's in operator, which offers an efficient way to determine if one string is a substring of another.

• Runtime Complexity: O(n^2 * m), where n is the number of words and m is the average length of the words.

• Storage Complexity: O(k), where k is the number of substring words.

Code

    def stringMatching(words):
    result = []
    for i in range(len(words)):
        for j in range(len(words)):
            if i != j and words[i] in words[j]:
                if words[i] not in result:
                    result.append(words[i])
    return result