Python: Check whether a number is "happy" or not
Python Basic - 1: Exercise-66 with Solution
From Wikipedia, the free encyclopaedia:
A happy number is defined by the following process:
Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers, while those that do not end in 1 are unhappy numbers.
Write a Python program to check whether a number is "happy" or not.
Sample Solution:
Python Code:
def is_Happy_num(n):
past = set()
while n != 1:
n = sum(int(i)**2 for i in str(n))
if n in past:
return False
past.add(n)
return True
print(is_Happy_num(7))
print(is_Happy_num(932))
print(is_Happy_num(6))
Sample Output:
True True False
Pictorial Presentation:
Flowchart:

Python Code Editor:
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Previous: Write a Python program to find the longest word in set of words which is a subsequence of a given string.
Next: Write a Python program to find and print the first 10 happy numbers.
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Python: Tips of the Day
How to make a flat list out of list of lists?
Given a list of lists l
flat_list = [item for sublist in l for item in sublist]
which means:
flat_list = [] for sublist in l: for item in sublist: flat_list.append(item)
is faster than the shortcuts posted so far. (l is the list to flatten.) Here is the corresponding function:
flatten = lambda l: [item for sublist in l for item in sublist]
As evidence, you can use the timeit module in the standard library:
$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' '[item for sublist in l for item in sublist]' 10000 loops, best of 3: 143 usec per loop $ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'sum(l, [])' 1000 loops, best of 3: 969 usec per loop $ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'reduce(lambda x,y: x+y,l)' 1000 loops, best of 3: 1.1 msec per loop
Explanation: the shortcuts based on + (including the implied use in sum) are, of necessity, O(L**2) when there are L sublists -- as the intermediate result list keeps getting longer, at each step a new intermediate result list object gets allocated, and all the items in the previous intermediate result must be copied over (as well as a few new ones added at the end). So, for simplicity and without actual loss of generality, say you have L sublists of I items each: the first I items are copied back and forth L-1 times, the second I items L-2 times, and so on; total number of copies is I times the sum of x for x from 1 to L excluded, i.e., I * (L**2)/2.
The list comprehension just generates one list, once, and copies each item over (from its original place of residence to the result list) also exactly once.
Ref: https://bit.ly/3dKsNTR
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