Introduction

In this blog post, we will explore how to solve the LeetCode problem "1291" 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

An integer has sequential digits if and only if each digit in the number is one more than the previous digit. Return a sorted list of all the integers in the range [low, high] inclusive that have sequential digits. Example 1: Input: low = 100, high = 300 Output: [123,234] Example 2: Input: low = 1000, high = 13000 Output: [1234,2345,3456,4567,5678,6789,12345] Constraints: 10 <= low <= high <= 10^9

Explanation

Here's the solution:

• Generate Sequential Numbers: Generate all possible sequential digit numbers starting from lengths 2 up to 9 (since the maximum possible number is 10^9).
• Filter by Range: Filter the generated sequential numbers to include only those within the given low and high range.

• Sort and Return: Sort the filtered numbers and return them as a list.

• Runtime Complexity: O(1), Storage Complexity: O(1) - because the number of sequential digits numbers is limited and doesn't scale with the input range.

Code

    def sequentialDigits(low: int, high: int) -> list[int]:
    result = []
    for start_digit in range(1, 10):
        num = start_digit
        for next_digit in range(start_digit + 1, 10):
            num = num * 10 + next_digit
            if low <= num <= high:
                result.append(num)
    result.sort()
    return result