# Solving Leetcode Interviews in Seconds with AI: Counting Words With a Given Prefix


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2185" 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
	> You are given an array of strings words and a string pref. Return the number of strings in words that contain pref as a prefix. A prefix of a string s is any leading contiguous substring of s.   Example 1:  Input: words = ["pay","attention","practice","attend"], pref = "at" Output: 2 Explanation: The 2 strings that contain "at" as a prefix are: "attention" and "attend".  Example 2:  Input: words = ["leetcode","win","loops","success"], pref = "code" Output: 0 Explanation: There are no strings that contain "code" as a prefix.    Constraints:  1 <= words.length <= 100 1 <= words[i].length, pref.length <= 100 words[i] and pref consist of lowercase English letters.  

	# Explanation
	*   **Iterate Through Words:** Loop through each word in the input array `words`.
*   **Check Prefix:** For each word, efficiently check if the input `pref` is a prefix of that word using string slicing.
*   **Count Matches:** Increment a counter if the prefix matches.

*   **Time Complexity:** O(N*M), where N is the number of words and M is the maximum length of a word or the prefix. **Space Complexity:** O(1).

	
	# Code
	```python
	def prefixCount(words: list[str], pref: str) -> int:
    """
    Counts the number of strings in 'words' that have 'pref' as a prefix.
    """
    count = 0
    for word in words:
        if word.startswith(pref):
            count += 1
    return count
	```