Introduction

In this blog post, we will explore how to solve the LeetCode problem "869" 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

You are given an integer n. We reorder the digits in any order (including the original order) such that the leading digit is not zero. Return true if and only if we can do this so that the resulting number is a power of two. Example 1: Input: n = 1 Output: true Example 2: Input: n = 10 Output: false Constraints: 1 <= n <= 109

Explanation

Here's the breakdown of the solution:

• Digit Counting: Convert the input integer n into a digit count representation (e.g., '121' becomes {1: 2, 2: 1}).
• Power of Two Iteration: Iterate through powers of two within a reasonable range (up to 2³¹, since n <= 10⁹). For each power of two, generate its digit count representation.

• Comparison: Compare the digit counts of the input number and each power of two. If a match is found, it means a rearrangement exists to form that power of two, so return True.

• Runtime Complexity: O(log(n) log(log(n))) where n is the upper bound. This is because we iterate through powers of 2 which takes O(log(n)) time and then compute the digit count which takes O(log(power_of_2)) = O(log(log(n))). *Storage Complexity: O(log(n)) due to the digit count dictionary.

Code

    from collections import Counter

def reordered_power_of_2(n: int) -> bool:
    """
    Determines if a reordering of the digits of n can form a power of two.
    """

    def digit_count(num: int) -> Counter:
        """
        Computes the digit count of a number.
        """
        counts = Counter()
        for digit in str(num):
            counts[digit] += 1
        return counts

    n_counts = digit_count(n)

    power = 1
    while power <= 10**9:
        if digit_count(power) == n_counts:
            return True
        power *= 2
    return False