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Update: Actually Wikipedia provides a clear argument that unhappy numbers will arrive in a loop containing the value 4, so this approach is valid, but should included a comment with a link to that proof.
def is_happy(n):
s = { 1 }
while n not in s:
s.add(n)
n = sum(i * i for i in map(int, str(n)))
return n == 1
Update:
Since Wikipedea has proof that all positive unhappy numbers end in the sequence 4 → 16 → 37 → 58 → 89 → 145 → 42 → 20 → 4 → ..., and happy numbers end in the sequence 1 → 1 → ..., you can create a set of these termination values (including 0 → 0 → ...), and no longer needed to maintain the set of "seen" values. By using all numbers in the unhappy loop, we can terminate the search up to 8 iterations earlier over just checking for n == 1 and n == 4.
def is_happy(num):
# See https://en.wikipedia.org/wiki/Happy_number#Sequence_behavior
terminal = { 0, 1, 4, 16, 20, 37, 42, 58, 89, 145 }
while num not in terminal:
num = sum(i * i for i in map(int, str(num)))
return n == 1
def is_happy(n):
s = { 1 }
while n not in s:
s.add(n)
n = sum(i * i for i in map(int, str(n)))
return n == 1
Update: Actually Wikipedia provides a clear argument that unhappy numbers will arrive in a loop containing the value 4, so this approach is valid, but should included a comment with a link to that proof.
def is_happy(n):
s = { 1 }
while n not in s:
s.add(n)
n = sum(i * i for i in map(int, str(n)))
return n == 1
Update:
Since Wikipedea has proof that all positive unhappy numbers end in the sequence 4 → 16 → 37 → 58 → 89 → 145 → 42 → 20 → 4 → ..., and happy numbers end in the sequence 1 → 1 → ..., you can create a set of these termination values (including 0 → 0 → ...), and no longer needed to maintain the set of "seen" values. By using all numbers in the unhappy loop, we can terminate the search up to 8 iterations earlier over just checking for n == 1 and n == 4.
def is_happy(num):
# See https://en.wikipedia.org/wiki/Happy_number#Sequence_behavior
terminal = { 0, 1, 4, 16, 20, 37, 42, 58, 89, 145 }
while num not in terminal:
num = sum(i * i for i in map(int, str(num)))
return n == 1
Solution #2:
class Solution2:
def isHappy(self, n):
# ...
while n != 1:
if n in s: return False
# ...
return n == 1
You are looping while n != 1, without any break statements. There is no need to test n == 1 at the return statement at the end. Just return True.
Solution #3 returns None if 0 is given as input, instead of returning True or False.
Solution #4 becomes an endless loop if 0 is given as input.
Are there any other stopping conditions other that n == 0, n == 1 or n == 4? It isn't clear that all unhappy numbers result in a loop containing the value 4, so the validity of this approach is in question.
In all your solutions, your loop is testing at least two conditions, such as both n != 1 and n is s. Why not initialize s to contain a 1 (or even just leave it as an empty set), and then only test n in s. No special cases.
def is_happy(n):
s = { 1 }
while n not in s:
s.add(n)
n = sum(i * i for i in map(int, str(n)))
return n == 1
Finally:
- follow the PEP-8 standards (avoid mixedCase function/method names),
- Stop writing classes!