Introduction

In this blog post, we will explore how to solve the LeetCode problem "1012" 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 number of positive integers in the range [1, n] that have at least one repeated digit. Example 1: Input: n = 20 Output: 1 Explanation: The only positive number (<= 20) with at least 1 repeated digit is 11. Example 2: Input: n = 100 Output: 10 Explanation: The positive numbers (<= 100) with atleast 1 repeated digit are 11, 22, 33, 44, 55, 66, 77, 88, 99, and 100. Example 3: Input: n = 1000 Output: 262 Constraints: 1 <= n <= 109

Explanation

Here's the breakdown of the solution:

• Calculate Numbers with Unique Digits: Instead of directly counting numbers with repeated digits, we calculate the count of numbers with unique digits within the range [1, n]. Then, we subtract this count from n to get the desired result.
• Digit-by-Digit Construction: We decompose n into its digits and construct numbers with unique digits from left to right, considering the constraints imposed by the digits of n. We use combinatorics to efficiently calculate the number of possibilities for each digit.

• Handling Leading Zeros: We need to be careful not to include numbers with leading zeros in our count of numbers with unique digits.

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

Code

    def num_with_repeated_digits(n: int) -> int:
    """
    Given an integer n, return the number of positive integers in the range [1, n]
    that have at least one repeated digit.

    For example:
    num_with_repeated_digits(20) == 1
    num_with_repeated_digits(100) == 10
    num_with_repeated_digits(1000) == 262
    """

    def count_unique_digits(num: int) -> int:
        s = str(num)
        length = len(s)
        ans = 0

        # Count numbers with length < len(s)
        for i in range(1, length):
            ans += 9 * perm(9, i - 1)

        # Count numbers with length == len(s)
        seen = set()
        for i in range(length):
            digit = int(s[i])
            for j in range(0 if i > 0 else 1, digit):
                if j not in seen:
                    ans += perm(10 - i - 1, length - i - 1)

            if digit in seen:
                break
            seen.add(digit)
        else:  # No repeated digits, include n
            ans += 1

        return ans

    def perm(n: int, k: int) -> int:
        res = 1
        for i in range(k):
            res *= (n - i)
        return res

    return n - count_unique_digits(n)