Leetcode - Power of Three (Python)

April 2021 Leetcode Challenge
Leetcode - Power of Three #326
Difficulty: Easy

[karthikkumaresan]

divisible is not enough.. if the quotient is 2 or 5 for example, then it is not going to work

[siwence]

don't know if you have notifs or not on but here's how the last code you showed works

since the max is 32 bit integer limit (2^31 -1 / 2147483647), they found the highest possible value were 3^x < 2^31 -1 (3^x because we're finding powers of 3)

in this case, it's 19.

since no number can be higher than that, you can check if the number is a power of 3 when 3^19 % 3 = 0 (no remainder)

therefore, it returns true if n will have no remainder

and false if it does have one

[ishan]

You're amazing mate! Thank you so much!