# Solving Leetcode Interviews in Seconds with AI: Number of Digit One


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "233" 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
	> Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.   Example 1:  Input: n = 13 Output: 6  Example 2:  Input: n = 0 Output: 0    Constraints:  0 <= n <= 109  

	# Explanation
	Here's the approach, complexity analysis, and Python code:

*   **Decomposition and Iteration:** Iterate through each digit place value (ones, tens, hundreds, etc.) from right to left. For each place value, determine how many times the digit '1' appears in that position across all numbers from 0 to `n`.
*   **Case Analysis:** For each digit position, analyze three cases: the digit at that position in `n` is 0, 1, or greater than 1. The number of '1's appearing in that position can be calculated based on these cases.
*   **Mathematical Formula:** Use a formula derived from the case analysis to calculate the number of '1's for each position and accumulate the result.

*   **Runtime Complexity:** O(log n) where n is the input number.
*   **Storage Complexity:** O(1)

	
	# Code
	```python
	def countDigitOne(n: int) -> int:
    """
    Counts the total number of digit 1 appearing in all non-negative integers less than or equal to n.

    Args:
        n: An integer.

    Returns:
        The total number of digit 1 appearing in all non-negative integers less than or equal to n.
    """
    count = 0
    factor = 1
    while factor <= n:
        digit = (n // factor) % 10
        remaining = n // (factor * 10)
        if digit == 0:
            count += remaining * factor
        elif digit == 1:
            count += remaining * factor + (n % factor) + 1
        else:
            count += (remaining + 1) * factor
        factor *= 10
    return count
	```