Understanding the Problem

Given a number a as a parameter, each digit should be squared then sum every squared digits. Check if the sum of the squared digits is 1. If it has a value of 1, return True, if it loops infinitely, return False.

Approach

First I divided the digits into a list then squared each digit. I needed to check if this process is looping infinitely with a temporary memory as a list.

Solution

class Solution:
    def isHappy(self, n: int) -> bool:
        num = n

        dup_list = []
        while True:
            calc_num = sum([int(i)**2 for i in str(num)])
            if calc_num == 1:
                return True
            elif calc_num != 1 and calc_num not in dup_list:
                dup_list.append(calc_num)
                num = calc_num
            else:
                return False

It was very simple and the instruction was clear. I calculated the summed number then put the value into calc_num. if the calc_num is a value of 1, I return True. If calc_num is not equal to 1 and it has not been shown yet during the loop, I keep this number in dup_list to check if this number appears again. If it appears again, I return False.

Reflection

Easy peasy(?) or pissie or whatever. At this point I feel like I have come a little bit further with my coding skills and this kind of a problem looks easier than before, like before I began this blog. I was thinking about breaking the loop first to be safe but since it returns True and False anyways, I decided not to.