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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:

Python: Check whether a number is 'happy' or not

Flowchart:

Flowchart: Python - Check whether a number is 'happy' or not

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