# Solving Leetcode Interviews in Seconds with AI: Minimum Number of Valid Strings to Form Target II


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3292" 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 target. A string x is called valid if x is a prefix of any string in words. Return the minimum number of valid strings that can be concatenated to form target. If it is not possible to form target, return -1.   Example 1:  Input: words = ["abc","aaaaa","bcdef"], target = "aabcdabc" Output: 3 Explanation: The target string can be formed by concatenating:  Prefix of length 2 of words[1], i.e. "aa". Prefix of length 3 of words[2], i.e. "bcd". Prefix of length 3 of words[0], i.e. "abc".   Example 2:  Input: words = ["abababab","ab"], target = "ababaababa" Output: 2 Explanation: The target string can be formed by concatenating:  Prefix of length 5 of words[0], i.e. "ababa". Prefix of length 5 of words[0], i.e. "ababa".   Example 3:  Input: words = ["abcdef"], target = "xyz" Output: -1    Constraints:  1 <= words.length <= 100 1 <= words[i].length <= 5 * 104 The input is generated such that sum(words[i].length) <= 105. words[i] consists only of lowercase English letters. 1 <= target.length <= 5 * 104 target consists only of lowercase English letters.  

	# Explanation
	Here's a breakdown of the approach, followed by the code:

*   **Dynamic Programming:** Use dynamic programming to store the minimum number of valid prefixes needed to form each prefix of the target string.
*   **Prefix Check:** Iterate through the `words` array and check if any prefix of a word matches a portion of the target string ending at the current index.
*   **Optimization:** Memoize the results to avoid redundant calculations and efficiently build the solution.

*   **Runtime Complexity:** O(m * n * k), where 'm' is the length of target, 'n' is the number of words, and 'k' is the average length of the words. **Storage Complexity:** O(m), where 'm' is the length of the target.

	
	# Code
	```python
	def min_prefixes(words, target):
    """
    Calculates the minimum number of valid prefixes needed to form the target string.

    Args:
      words: A list of strings.
      target: The target string.

    Returns:
      The minimum number of valid prefixes, or -1 if the target cannot be formed.
    """

    n = len(target)
    dp = [float('inf')] * (n + 1)
    dp[0] = 0

    for i in range(1, n + 1):
        for word in words:
            for j in range(1, len(word) + 1):
                if i >= j and target[i-j:i] == word[:j]:
                    dp[i] = min(dp[i], dp[i-j] + 1)

    if dp[n] == float('inf'):
        return -1
    else:
        return dp[n]
	```