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Python: Add two positive integers without using the '+' operator

Python Basic - 1: Exercise-14 with Solution

Write a Python program to add two positive integers without using the '+' operator.

Note: Use bitwise operations to add two numbers.

Sample Solution:

Python Code:

def add_without_plus_operator(a, b):
    while b != 0:
        data = a & b
        a = a ^ b
        b = data << 1
    return a
print(add_without_plus_operator(2, 10))
print(add_without_plus_operator(-20, 10))
print(add_without_plus_operator(-10, -20))

Sample Output:

12
-10
-30

Pictorial Presentation:

Python Exercises: Add two positive integers without using the '+' operator.

Flowchart:

Flowchart: Python - Add two positive integers without using the '+' operator

Visualize Python code execution:

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Python Code Editor :

 

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Previous: Write a Python program to get all possible two digit letter combinations from a digit (1 to 9) string.
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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