Introduction

In this blog post, we will explore how to solve the LeetCode problem "2376" 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

We call a positive integer special if all of its digits are distinct. Given a positive integer n, return the number of special integers that belong to the interval [1, n]. Example 1: Input: n = 20 Output: 19 Explanation: All the integers from 1 to 20, except 11, are special. Thus, there are 19 special integers. Example 2: Input: n = 5 Output: 5 Explanation: All the integers from 1 to 5 are special. Example 3: Input: n = 135 Output: 110 Explanation: There are 110 integers from 1 to 135 that are special. Some of the integers that are not special are: 22, 114, and 131. Constraints: 1 <= n <= 2 * 109

Explanation

Here's the breakdown:

• Decomposition: Break down the problem into counting special numbers with fewer digits than n, and then count special numbers with the same number of digits as n, considering the prefix digits.
• Dynamic Programming/Combinatorics: Use combinatorics to count special numbers with fewer digits. Use dynamic programming principles to avoid redundant calculations when counting special numbers with the same number of digits.

• Digit-by-Digit Analysis: Process the digits of n one by one, keeping track of the digits used so far. This allows us to efficiently determine the count of special numbers that are less than or equal to the current prefix of n.

• Runtime Complexity: O(log n), where n is the input integer. This is because the number of digits in n is logarithmic with respect to n. Storage Complexity: O(log n) to store the digits of n.

Code

    def countSpecialNumbers(n: int) -> int:
    """
    Counts the number of special integers in the range [1, n].

    A positive integer is special if all of its digits are distinct.
    """

    s = str(n)
    k = len(s)
    ans = 0

    # Count special numbers with fewer digits
    for i in range(1, k):
        ans += 9 * factorial(9) // factorial(10 - i)

    # Count special numbers with the same number of digits
    used = set()
    for i, digit in enumerate(s):
        d = int(digit)
        for j in range(0 if i > 0 else 1, d):
            if j not in used:
                ans += factorial(9 - i) // factorial(10 - k)

        if d in used:
            break
        used.add(d)
        if i == k - 1:
            ans += 1  # n itself is special

    return ans

def factorial(n):
    """Calculates the factorial of n."""
    res = 1
    for i in range(1, n + 1):
        res *= i
    return res