# Solving Leetcode Interviews in Seconds with AI: Count Prefix and Suffix Pairs I


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3042" 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 a 0-indexed string array words. Let's define a boolean function isPrefixAndSuffix that takes two strings, str1 and str2:  isPrefixAndSuffix(str1, str2) returns true if str1 is both a prefix and a suffix of str2, and false otherwise.  For example, isPrefixAndSuffix("aba", "ababa") is true because "aba" is a prefix of "ababa" and also a suffix, but isPrefixAndSuffix("abc", "abcd") is false. Return an integer denoting the number of index pairs (i, j) such that i < j, and isPrefixAndSuffix(words[i], words[j]) is true.   Example 1:  Input: words = ["a","aba","ababa","aa"] Output: 4 Explanation: In this example, the counted index pairs are: i = 0 and j = 1 because isPrefixAndSuffix("a", "aba") is true. i = 0 and j = 2 because isPrefixAndSuffix("a", "ababa") is true. i = 0 and j = 3 because isPrefixAndSuffix("a", "aa") is true. i = 1 and j = 2 because isPrefixAndSuffix("aba", "ababa") is true. Therefore, the answer is 4. Example 2:  Input: words = ["pa","papa","ma","mama"] Output: 2 Explanation: In this example, the counted index pairs are: i = 0 and j = 1 because isPrefixAndSuffix("pa", "papa") is true. i = 2 and j = 3 because isPrefixAndSuffix("ma", "mama") is true. Therefore, the answer is 2.   Example 3:  Input: words = ["abab","ab"] Output: 0 Explanation: In this example, the only valid index pair is i = 0 and j = 1, and isPrefixAndSuffix("abab", "ab") is false. Therefore, the answer is 0.   Constraints:  1 <= words.length <= 50 1 <= words[i].length <= 10 words[i] consists only of lowercase English letters.  

	# Explanation
	Here's the solution:

*   **High-level approach:** The solution iterates through all possible pairs of indices (i, j) such that i < j. For each pair, it checks if `words[i]` is both a prefix and a suffix of `words[j]` using a helper function. If it is, the count is incremented.

*   **Complexity:** Time Complexity: O(n^2 * m), where n is the number of words and m is the maximum length of a word. Space Complexity: O(1).

	
	# Code
	```python
	def count_prefix_suffix_pairs(words):
    """
    Counts the number of index pairs (i, j) such that i < j, and isPrefixAndSuffix(words[i], words[j]) is true.
    """

    def is_prefix_and_suffix(str1, str2):
        """
        Checks if str1 is both a prefix and a suffix of str2.
        """
        n1 = len(str1)
        n2 = len(str2)

        if n1 > n2:
            return False

        # Check if str1 is a prefix of str2
        if str2[:n1] != str1:
            return False

        # Check if str1 is a suffix of str2
        if str2[n2 - n1:] != str1:
            return False

        return True

    count = 0
    n = len(words)
    for i in range(n):
        for j in range(i + 1, n):
            if is_prefix_and_suffix(words[i], words[j]):
                count += 1

    return count
	```