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, 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

    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]