Introduction

In this blog post, we will explore how to solve the LeetCode problem "139" 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

Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a space-separated sequence of one or more dictionary words. Note that the same word in the dictionary may be reused multiple times in the segmentation. Example 1: Input: s = "leetcode", wordDict = ["leet","code"] Output: true Explanation: Return true because "leetcode" can be segmented as "leet code". Example 2: Input: s = "applepenapple", wordDict = ["apple","pen"] Output: true Explanation: Return true because "applepenapple" can be segmented as "apple pen apple". Note that you are allowed to reuse a dictionary word. Example 3: Input: s = "catsandog", wordDict = ["cats","dog","sand","and","cat"] Output: false Constraints: 1 <= s.length <= 300 1 <= wordDict.length <= 1000 1 <= wordDict[i].length <= 20 s and wordDict[i] consist of only lowercase English letters. All the strings of wordDict are unique.

Explanation

Here's the solution to the word break problem:

• Dynamic Programming: Use a boolean array dp where dp[i] represents whether the substring s[0:i] can be segmented.
• Iteration: Iterate through the string s and for each index i, check if any word in wordDict can form the suffix s[j:i] and if dp[j] is true.

• Base Case: dp[0] is initialized to True because an empty string can always be segmented.

• Runtime Complexity: O(n*m*k) where n is the length of s, m is the number of words in wordDict, and k is the average length of a word in wordDict. Storage Complexity: O(n)

Code

    def wordBreak(s: str, wordDict: list[str]) -> bool:
    """
    Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a space-separated sequence of one or more dictionary words.

    Args:
        s (str): The input string.
        wordDict (list[str]): The dictionary of words.

    Returns:
        bool: True if s can be segmented, False otherwise.
    """
    n = len(s)
    dp = [False] * (n + 1)
    dp[0] = True

    for i in range(1, n + 1):
        for word in wordDict:
            if i >= len(word) and s[i - len(word):i] == word:
                dp[i] = dp[i] or dp[i - len(word)]

    return dp[n]