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


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3291" 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 * 103 The input is generated such that sum(words[i].length) <= 105. words[i] consists only of lowercase English letters. 1 <= target.length <= 5 * 103 target consists only of lowercase English letters.  

	# Explanation
	Here's the solution:

*   **Dynamic Programming:** Use dynamic programming to determine the minimum number of valid string concatenations to form the target string up to a given index. `dp[i]` stores the minimum number of valid strings to form `target[:i]`.
*   **Prefix Check:** For each index in the target string, iterate through the `words` to check if any word's prefix matches the substring of the target ending at that index.
*   **Base Case and Transition:** Initialize `dp[0]` to 0 (empty string requires 0 valid strings). If a valid prefix is found, update `dp[i]` using `dp[i] = min(dp[i], dp[j] + 1)`, where `j` is the starting index of the prefix in target.

*   **Runtime Complexity:** O(M\*N\*L), where M is the length of target, N is the number of words, and L is the maximum length of a word.
*   **Storage Complexity:** O(M), where M is the length of the target string.

	
	# Code
	```python
	def min_concatenations(words, target):
    n = len(target)
    dp = [float('inf')] * (n + 1)
    dp[0] = 0

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