Introduction

In this blog post, we will explore how to solve the LeetCode problem "902" 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 array of digits which is sorted in non-decreasing order. You can write numbers using each digits[i] as many times as we want. For example, if digits = ['1','3','5'], we may write numbers such as '13', '551', and '1351315'. Return the number of positive integers that can be generated that are less than or equal to a given integer n. Example 1: Input: digits = ["1","3","5","7"], n = 100 Output: 20 Explanation: The 20 numbers that can be written are: 1, 3, 5, 7, 11, 13, 15, 17, 31, 33, 35, 37, 51, 53, 55, 57, 71, 73, 75, 77. Example 2: Input: digits = ["1","4","9"], n = 1000000000 Output: 29523 Explanation: We can write 3 one digit numbers, 9 two digit numbers, 27 three digit numbers, 81 four digit numbers, 243 five digit numbers, 729 six digit numbers, 2187 seven digit numbers, 6561 eight digit numbers, and 19683 nine digit numbers. In total, this is 29523 integers that can be written using the digits array. Example 3: Input: digits = ["7"], n = 8 Output: 1 Constraints: 1 <= digits.length <= 9 digits[i].length == 1 digits[i] is a digit from '1' to '9'. All the values in digits are unique. digits is sorted in non-decreasing order. 1 <= n <= 109

Explanation

• Count numbers shorter than n: Calculate the number of possible integers with lengths less than the length of n. This is a straightforward power calculation based on the number of available digits.
  • Count numbers of the same length as n: Iterate through the digits of n and, for each digit, count how many digits in the digits array are less than it. Recursively continue the process for the remaining digits of n.
  • Handle the case where a prefix of n can be exactly formed: If at any point, a digit in digits is equal to the corresponding digit of n, it means we can potentially form n itself or other numbers with the same prefix.

• Runtime Complexity: O(log n), where n is the input integer.
• Storage Complexity: O(log n), mainly due to the recursion stack.

Code

    def atMostNGivenDigitSet(digits, n):
    s = str(n)
    k = len(s)
    ans = 0

    # Count numbers shorter than n
    for i in range(1, k):
        ans += len(digits) ** i

    # Count numbers of the same length as n
    for i, digit in enumerate(s):
        count = 0
        for d in digits:
            if d < digit:
                count += 1
            elif d == digit:
                ans += count
                if i == k - 1:
                    ans += 1
                break
            else:
                ans += count
                return ans
        else:
            return ans

    return ans