# Solving Leetcode Interviews in Seconds with AI: Check if Number is a Sum of Powers of Three


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1780" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), 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 possible to represent n as the sum of distinct powers of three. Otherwise, return false. An integer y is a power of three if there exists an integer x such that y == 3x.   Example 1:  Input: n = 12 Output: true Explanation: 12 = 31 + 32  Example 2:  Input: n = 91 Output: true Explanation: 91 = 30 + 32 + 34  Example 3:  Input: n = 21 Output: false    Constraints:  1 <= n <= 107  

	# Explanation
	Here's the breakdown of the solution:

*   **Base 3 Representation:** Convert the given number `n` into its base-3 representation. If `n` can be represented as a sum of distinct powers of three, its base-3 representation will only contain the digits 0 and 1.

*   **Check for '2' Digit:** Iterate through the digits of the base-3 representation. If any digit is equal to 2, it means that the corresponding power of three appears twice in the sum, violating the "distinct powers" constraint.

*   **Return Value:** Return `True` if the base-3 representation contains only 0s and 1s, and `False` otherwise.

*   **Time & Space Complexity:** O(log3(n)) time, O(1) space (dominant factor is converting to base 3 and iterating through the digits, and the space used is constant).

	
	# Code
	```python
	def check_if_sum_of_distinct_powers(n: int) -> bool:
    """
    Given an integer n, return true if it is possible to represent n as the sum of distinct powers of three.
    Otherwise, return false.
    """
    while n > 0:
        if n % 3 == 2:
            return False
        n //= 3
    return True
	```