Introduction

In this blog post, we will explore how to solve the LeetCode problem "1849" 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 string s that consists of only digits. Check if we can split s into two or more non-empty substrings such that the numerical values of the substrings are in descending order and the difference between numerical values of every two adjacent substrings is equal to 1. For example, the string s = "0090089" can be split into ["0090", "089"] with numerical values [90,89]. The values are in descending order and adjacent values differ by 1, so this way is valid. Another example, the string s = "001" can be split into ["0", "01"], ["00", "1"], or ["0", "0", "1"]. However all the ways are invalid because they have numerical values [0,1], [0,1], and [0,0,1] respectively, all of which are not in descending order. Return true if it is possible to split s​​​​​​ as described above, or false otherwise. A substring is a contiguous sequence of characters in a string. Example 1: Input: s = "1234" Output: false Explanation: There is no valid way to split s. Example 2: Input: s = "050043" Output: true Explanation: s can be split into ["05", "004", "3"] with numerical values [5,4,3]. The values are in descending order with adjacent values differing by 1. Example 3: Input: s = "9080701" Output: false Explanation: There is no valid way to split s. Constraints: 1 <= s.length <= 20 s only consists of digits.

Explanation

Here's the solution:

• High-level Approach:

  • Iterate through all possible starting lengths of the first substring.
  • For each starting length, check if the remaining string can be split into substrings that satisfy the descending order and difference of 1 condition. This is done recursively or iteratively.
  • If a valid split is found, return True. Otherwise, continue with the next starting length.

• Complexity:

  • Runtime Complexity: O(N^2), where N is the length of the input string. This is because we iterate through all possible starting lengths for the first substring (O(N)), and for each starting length, the check_split function potentially iterates through the remaining string in O(N) in the worst case.
  • Storage Complexity: O(N) due to recursion depth in the check_split function, at most. If we implement this iteratively, the storage complexity is O(1).

Code

    def split_string(s: str) -> bool:
    """
    Checks if a string can be split into substrings such that the numerical
    values of the substrings are in descending order and the difference
    between numerical values of every two adjacent substrings is equal to 1.
    """

    def check_split(s: str, prev_val: int) -> bool:
        """
        Recursively checks if the remaining string can be split to satisfy
        the conditions given the value of the previous substring.
        """
        if not s:
            return True

        for i in range(1, len(s) + 1):
            curr_str = s[:i]
            curr_val = int(curr_str)
            if curr_val == prev_val - 1:
                if check_split(s[i:], curr_val):
                    return True
        return False

    for i in range(1, len(s) + 1):
        first_str = s[:i]
        first_val = int(first_str)
        if check_split(s[i:], first_val):
            return True

    return False