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Solving Leetcode Interviews in Seconds with AI: Power of Four

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2 min read

Introduction

In this blog post, we will explore how to solve the LeetCode problem "342" 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 true if it is a power of four. Otherwise, return false. An integer n is a power of four, if there exists an integer x such that n == 4x. Example 1: Input: n = 16 Output: true Example 2: Input: n = 5 Output: false Example 3: Input: n = 1 Output: true Constraints: -231 <= n <= 231 - 1 Follow up: Could you solve it without loops/recursion?

Explanation

Here's the breakdown of the approach and the Python code:

  • High-Level Approach:

    • Check if the number is a power of 2. If not, it cannot be a power of 4.
    • If it's a power of 2, check if the only set bit is at an even position. This can be done efficiently using a bitmask. Powers of 4 in binary representation have the form 1, 100, 10000, etc. (only a single '1' bit at an even position).
  • Complexity:

    • Runtime Complexity: O(1)
    • Storage Complexity: O(1)

Code

    def isPowerOfFour(n: int) -> bool:
    """
    Given an integer n, return true if it is a power of four. Otherwise, return false.
    An integer n is a power of four, if there exists an integer x such that n == 4x.
    """
    if n <= 0:
        return False
    if (n & (n - 1)) != 0:  # Check if it's a power of 2
        return False
    return (n & 0x55555555) != 0  # Check if the set bit is at an even position

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