# Solving Leetcode Interviews in Seconds with AI: Sequential Digits


	# 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](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
	> 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
	```python
	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
	```