Introduction

In this blog post, we will explore how to solve the LeetCode problem "264" 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

An ugly number is a positive integer whose prime factors are limited to 2, 3, and 5. Given an integer n, return the nth ugly number. Example 1: Input: n = 10 Output: 12 Explanation: [1, 2, 3, 4, 5, 6, 8, 9, 10, 12] is the sequence of the first 10 ugly numbers. Example 2: Input: n = 1 Output: 1 Explanation: 1 has no prime factors, therefore all of its prime factors are limited to 2, 3, and 5. Constraints: 1 <= n <= 1690

Explanation

Here's an efficient solution to find the nth ugly number:

• Dynamic Programming: Build an array of ugly numbers in increasing order. Each ugly number is a multiple of 2, 3, or 5 from a previous ugly number.
• Pointers: Maintain three pointers, one for each of the prime factors 2, 3, and 5. These pointers indicate the index of the smallest ugly number to multiply with each prime factor.

• Min Heap Alternative (Less Efficient): While a min-heap could work, the dynamic programming approach is generally faster due to less overhead.

• Time & Space Complexity: O(n) time, O(n) space

Code

    def nthUglyNumber(n: int) -> int:
    """
    Finds the nth ugly number.

    Args:
        n: The index of the ugly number to find (1-indexed).

    Returns:
        The nth ugly number.
    """
    if n <= 0:
        return 0

    ugly_numbers = [1] * n
    p2 = p3 = p5 = 0

    for i in range(1, n):
        ugly_numbers[i] = min(ugly_numbers[p2] * 2, ugly_numbers[p3] * 3, ugly_numbers[p5] * 5)

        if ugly_numbers[i] == ugly_numbers[p2] * 2:
            p2 += 1
        if ugly_numbers[i] == ugly_numbers[p3] * 3:
            p3 += 1
        if ugly_numbers[i] == ugly_numbers[p5] * 5:
            p5 += 1

    return ugly_numbers[n - 1]