Checking if a number is the sum of 2 other numbers

Question

Problem Statement

Given a list of numbers and a number k, return whether any two numbers from the list add up to k.

Example

Given [1, 2, 3] and k = 5, return True since 2 + 3 = 5.

This is what I've tried to do:

def pairs(n):
    for i in range(len(n)):
        for j in range(i+1, len()):
            yield n[i], n[j]


def ListCheck():
    number = input("Give me a number:")
    val = int(number)
    nums = [1,2,3]
    for i, j in pairs(nums):
        if j + i == val:
            print(True)
            break


ListCheck()

I'm getting an error when I run it, and I can't understand why.

what happens when you try to run it? an error? no error but unexpected out? can you specify what "doesn't work" means? — Paritosh Singh, Jan 01 '19 at 11:19
What error? there are different ones that tell different problems - please _always_ print the stacktrace of the error — Patrick Artner, Jan 01 '19 at 11:27

U13-Forward · Answer 1 · 2019-01-01T11:48:13.130

6

You could also do itertools.combinations, little shorter than @bitto's solution:

import itertools
def f(lst,num):
    for x,y in itertools.combinations(lst,2):
        if x+y==num:
            return True
    return False
lst=[1,2,3]
num=int(input("Give me a number: "))
print(f(lst,num))

edited Jan 01 '19 at 11:48

answered Jan 01 '19 at 11:38

U13-Forward

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This is the correct answer that does not provide wrong outputs – DeepSpace Jan 01 '19 at 11:46

Bitto · Accepted Answer · 2019-01-01T11:45:40.060

2

def issumoftwo(lst,num):
    for x in lst:
        for y in lst:
            if x+y==num and lst.index(x)!=lst.index(y):
                return True
    return False
lst=[1,2,3]
num=int(input("Give me a Number: "))
print(issumoftwo(lst,num))

Output

Give me a number: 5
True

edited Jan 01 '19 at 11:45

answered Jan 01 '19 at 11:19

Bitto

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I wan't to be able to break it when I find a solution. – Rony Kositsky Jan 01 '19 at 11:21
@RonyKositsky once you hit the return statement, you return from the function – Bitto Jan 01 '19 at 11:22
@RonyKositsky https://stackoverflow.com/questions/11604699/is-there-a-way-to-do-more-work-after-a-return-statement – Bitto Jan 01 '19 at 11:24
OK, now I understand. Thanks you! – Rony Kositsky Jan 01 '19 at 11:28
FWIW 1. that's `O(n^2)`, and there are several `O(n)` solutions. 2. This will sometimes wrongly return `True` (consider `list = [1]` and `num == 2`) – DeepSpace Jan 01 '19 at 11:29
@DeepSpace Is there any other case other than the one where len(list) is 1? I never thought about that to be frank. – Bitto Jan 01 '19 at 11:38
the problem was a syntax error on an example of yield. You provide an other approach. O(n) solutions need sets. – B. M. Jan 01 '19 at 11:40
@bitto Yes, almost every input can result in a wrong output, `[1, 2, 3]` and `6` – DeepSpace Jan 01 '19 at 11:41
@DeepSpace I should not have answered in haste. I think i have solved that particular problem for now. – Bitto Jan 01 '19 at 11:47

Patrick Artner · Answer 3 · 2019-01-01T11:55:42.120

You misses an n inside len(). The error

TypeError: len() takes exactly one argument (0 given)

tells you exactly what is wrong (if you fix the indentation problems of your code postet above).

You can streamline your code by using itertools.combinations. If you add some parameters to a function, you can generalize the problem searching as well - to get all combinations of n numbers from your list that add up to your targetvalue.

from itertools import combinations 

def is_sum_of_n_numbers(data ,target_value, num_elem):
    """Returns 'True' if any combinatin of 'num_elem'ents 
    from 'data' sums to 'target_value'"""
    return any(sum(x)==target_value for x in combinations(data, num_elem))

def find_sum_in_combination(data, target_value, num_elem):
    """Returns all combinations of 'num_elem'ent-tuples from 'data' 
    that sums to 'target_value'"""
    return [x for x in combinations(data,num_elem) if sum(x) == target_value]

Get all of them:

d = [1,2,3,4,5]
for numbers in range(1,6):
    for s in range(1,sum(d)+1):
        result = find_sum_in_combination(d,s,numbers)
        if result:
            print(f"Sum {s} from {d} with {numbers} numbers: ", result)

Output:

Sum 1 from [1, 2, 3, 4, 5] with 1 numbers:  [(1,)]
Sum 2 from [1, 2, 3, 4, 5] with 1 numbers:  [(2,)]
Sum 3 from [1, 2, 3, 4, 5] with 1 numbers:  [(3,)]
Sum 4 from [1, 2, 3, 4, 5] with 1 numbers:  [(4,)]
Sum 5 from [1, 2, 3, 4, 5] with 1 numbers:  [(5,)]
Sum 3 from [1, 2, 3, 4, 5] with 2 numbers:  [(1, 2)]
Sum 4 from [1, 2, 3, 4, 5] with 2 numbers:  [(1, 3)]
Sum 5 from [1, 2, 3, 4, 5] with 2 numbers:  [(1, 4), (2, 3)]
Sum 6 from [1, 2, 3, 4, 5] with 2 numbers:  [(1, 5), (2, 4)]
Sum 7 from [1, 2, 3, 4, 5] with 2 numbers:  [(2, 5), (3, 4)]
Sum 8 from [1, 2, 3, 4, 5] with 2 numbers:  [(3, 5)]
Sum 9 from [1, 2, 3, 4, 5] with 2 numbers:  [(4, 5)]
Sum 6 from [1, 2, 3, 4, 5] with 3 numbers:  [(1, 2, 3)]
Sum 7 from [1, 2, 3, 4, 5] with 3 numbers:  [(1, 2, 4)]
Sum 8 from [1, 2, 3, 4, 5] with 3 numbers:  [(1, 2, 5), (1, 3, 4)]
Sum 9 from [1, 2, 3, 4, 5] with 3 numbers:  [(1, 3, 5), (2, 3, 4)]
Sum 10 from [1, 2, 3, 4, 5] with 3 numbers:  [(1, 4, 5), (2, 3, 5)]
Sum 11 from [1, 2, 3, 4, 5] with 3 numbers:  [(2, 4, 5)]
Sum 12 from [1, 2, 3, 4, 5] with 3 numbers:  [(3, 4, 5)]
Sum 10 from [1, 2, 3, 4, 5] with 4 numbers:  [(1, 2, 3, 4)]
Sum 11 from [1, 2, 3, 4, 5] with 4 numbers:  [(1, 2, 3, 5)]
Sum 12 from [1, 2, 3, 4, 5] with 4 numbers:  [(1, 2, 4, 5)]
Sum 13 from [1, 2, 3, 4, 5] with 4 numbers:  [(1, 3, 4, 5)]
Sum 14 from [1, 2, 3, 4, 5] with 4 numbers:  [(2, 3, 4, 5)]
Sum 15 from [1, 2, 3, 4, 5] with 5 numbers:  [(1, 2, 3, 4, 5)]

Doku:

score 0 · Answer 4 · answered Jan 01 '19 at 11:32

The logic of your code is correct (It's unnecessarily complex as you can see by comparing with the above answer, but still correct ;)....)
Just check your indentations and put for j in range(i+1, len(n)) in the 3rd line of your code...you forgot the 'n'!! You need to give atleast one argument to len.

score 0 · Answer 5 · answered Jan 01 '19 at 11:35

0

You can solve your solution this way

n=[3,2,1]
number=int(input("Please enter nubmer"))
for i in n:
    num=number - i
    if num in n:
        print(num,i)
        break

This is solution but need to be customize as you want

answered Jan 01 '19 at 11:35

Mohammad Ansari

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This will return numbers also – Mohammad Ansari Jan 01 '19 at 11:36
This is one of `O(n)` solutions I was referring to, however still has the bug of returning `True` in case `list = [1] and num = 2`. – DeepSpace Jan 01 '19 at 11:36
It may return the wrong output, see my comments to @bitt's answer – DeepSpace Jan 01 '19 at 11:43
@DeepSpace Thanks for your help bro – Mohammad Ansari Jan 01 '19 at 15:36

score 0 · Answer 6 · answered Apr 03 '22 at 11:38

simple checking (k - arrayValue) will not work because if k = 4 and there is one 2 in array - it will be false positive
so you need to exclude cheked values from further loops
it can be done with pop function for arrays

python solution:

def ListCheck():
    number = 11
    nums = [10, 15, 3, 7, 9, 6, 4, 8]
    found = 0
    while found == 0:
        if len(nums) < 2: # check if array has at least 2 numbers
            break
        while len(nums) > 1:
            comparenum = nums.pop() # removes comparable number from array to avoid false true if nums has item == number*2 
            if (number - comparenum) in nums:
                print(True)
                found = 1
                break

    if found == 0:
        print(False)
        

ListCheck()
exit

Checking if a number is the sum of 2 other numbers

6 Answers6