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Python: Sum of the last digit of first number and the last digit of second number equal to the last digit of third number

Python Basic - 1: Exercise-108 with Solution

Write a Python program that takes three integers and check whether the sum of the last digit of first number and the last digit of second number equal to the last digit of third number.

Sample Solution:

Python Code:

def check_last_digit(x, y, z):
  return str(x+z)[-1] == str(y)[-1]
print(check_last_digit(12, 26, 44))
print(check_last_digit(145, 122, 1010))
print(check_last_digit(0, 20, 40))
print(check_last_digit(1, 22, 40))
print(check_last_digit(145, 129, 104))

Sample Output:

True
False
True
False
True

Pictorial Presentation:

Python: Sum of the last digit of first number and  the last digit of second number  equal to the last digit of third number.
Python: Sum of the last digit of first number and  the last digit of second number  equal to the last digit of third number.

Flowchart:

Flowchart: Python - Sum of the last digit of first number and  the last digit of second number  equal to the last digit of third number.

Python Code Editor:

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Previous: Write a Python program to check whether a given number is Oddish or Evenish.
Next: Write a Python program find the indices of all occurrences of a given item in a given list.

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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