Introduction

In this blog post, we will explore how to solve the LeetCode problem "357" 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 an integer n, return the count of all numbers with unique digits, x, where 0 <= x < 10n. Example 1: Input: n = 2 Output: 91 Explanation: The answer should be the total numbers in the range of 0 ≤ x < 100, excluding 11,22,33,44,55,66,77,88,99 Example 2: Input: n = 0 Output: 1 Constraints: 0 <= n <= 8

Explanation

Here's the solution to the problem, explained with high-level approach, complexity analysis, and the Python code.

• High-level approach:

  • Handle the base cases (n = 0 and n = 1) separately.
  • Iteratively calculate the count of numbers with unique digits for each length from 2 up to n. For each length, the count is derived by multiplying the number of available digits for each position. Handle leading zero carefully.
  • Sum up the counts for all lengths from 1 to n (inclusive), plus the base case count for n=0 if n>0 or n=0.

• Complexity Analysis:

  • Runtime Complexity: O(n), where n is the input integer.
  • Storage Complexity: O(1)

Code

    def countNumbersWithUniqueDigits(n: int) -> int:
    """
    Given an integer n, return the count of all numbers with unique digits, x, where 0 <= x < 10^n.

    For example:
    countNumbersWithUniqueDigits(2) == 91
    countNumbersWithUniqueDigits(0) == 1
    """

    if n == 0:
        return 1

    count = 10  # For n = 1 (0 to 9)
    unique_digits = 9
    available_number = 9

    for i in range(2, min(n + 1, 11)):
        unique_digits = unique_digits * available_number
        count += unique_digits
        available_number -= 1

    return count