# Solving Leetcode Interviews in Seconds with AI: Extra Characters in a String


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2707" 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 s and a dictionary of words dictionary. You have to break s into one or more non-overlapping substrings such that each substring is present in dictionary. There may be some extra characters in s which are not present in any of the substrings. Return the minimum number of extra characters left over if you break up s optimally.   Example 1:  Input: s = "leetscode", dictionary = ["leet","code","leetcode"] Output: 1 Explanation: We can break s in two substrings: "leet" from index 0 to 3 and "code" from index 5 to 8. There is only 1 unused character (at index 4), so we return 1.   Example 2:  Input: s = "sayhelloworld", dictionary = ["hello","world"] Output: 3 Explanation: We can break s in two substrings: "hello" from index 3 to 7 and "world" from index 8 to 12. The characters at indices 0, 1, 2 are not used in any substring and thus are considered as extra characters. Hence, we return 3.    Constraints:  1 <= s.length <= 50 1 <= dictionary.length <= 50 1 <= dictionary[i].length <= 50 dictionary[i] and s consists of only lowercase English letters dictionary contains distinct words  

	# Explanation
	Here's the breakdown of the solution and the Python code:

*   **High-Level Approach:**
    *   Dynamic Programming: Use DP to store the minimum extra characters for prefixes of the string `s`.
    *   Iterate through the string `s`: For each index `i`, calculate the minimum extra characters needed up to that index.
    *   Check Dictionary: For each index `i`, iterate through the `dictionary` to see if a word ends at `i`. If a word ends at `i`, update the DP table considering that word.

*   **Complexity:**
    *   Runtime Complexity: O(n\*m\*k) where n is the length of s, m is the length of dictionary, and k is the average length of words in dictionary.
    *   Storage Complexity: O(n)

	
	# Code
	```python
	def minExtraChar(s: str, dictionary: list[str]) -> int:
    """
    Given a 0-indexed string s and a dictionary of words dictionary, you have to break s into one or more non-overlapping substrings such that each substring is present in dictionary. There may be some extra characters in s which are not present in any of the substrings. Return the minimum number of extra characters left over if you break up s optimally.
    """
    n = len(s)
    dp = [0] * (n + 1)

    for i in range(1, n + 1):
        dp[i] = dp[i - 1] + 1  # Assume the current char is extra
        for word in dictionary:
            if i >= len(word) and s[i - len(word):i] == word:
                dp[i] = min(dp[i], dp[i - len(word)])

    return dp[n]
	```