# Solving Leetcode Interviews in Seconds with AI: Number of Strings That Appear as Substrings in Word


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1967" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), 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 strings patterns and a string word, return the number of strings in patterns that exist as a substring in word. A substring is a contiguous sequence of characters within a string.   Example 1:  Input: patterns = ["a","abc","bc","d"], word = "abc" Output: 3 Explanation: - "a" appears as a substring in "abc". - "abc" appears as a substring in "abc". - "bc" appears as a substring in "abc". - "d" does not appear as a substring in "abc". 3 of the strings in patterns appear as a substring in word.  Example 2:  Input: patterns = ["a","b","c"], word = "aaaaabbbbb" Output: 2 Explanation: - "a" appears as a substring in "aaaaabbbbb". - "b" appears as a substring in "aaaaabbbbb". - "c" does not appear as a substring in "aaaaabbbbb". 2 of the strings in patterns appear as a substring in word.  Example 3:  Input: patterns = ["a","a","a"], word = "ab" Output: 3 Explanation: Each of the patterns appears as a substring in word "ab".    Constraints:  1 <= patterns.length <= 100 1 <= patterns[i].length <= 100 1 <= word.length <= 100 patterns[i] and word consist of lowercase English letters.  

	# Explanation
	Here's a solution to the problem:

*   **Iterate through patterns:** The core idea is to loop through each string in the `patterns` array.
*   **Substring Check:** For each pattern, efficiently check if it's a substring of the given `word` using the `in` operator (which utilizes optimized substring search algorithms).
*   **Count Occurrences:** Increment a counter each time a pattern is found as a substring within the word.

*   **Runtime & Storage Complexity:** O(m\*n\*k), where n is the length of `patterns`, m is the length of `word`, and k is the average length of strings in `patterns`. Storage complexity is O(1).

	
	# Code
	```python
	def numOfStrings(patterns, word):
    """
    Given an array of strings patterns and a string word, return the number of strings in patterns that exist as a substring in word.

    Args:
        patterns (list[str]): A list of strings.
        word (str): A string.

    Returns:
        int: The number of strings in patterns that are substrings of word.
    """
    count = 0
    for pattern in patterns:
        if pattern in word:
            count += 1
    return count
	```