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.

Answer 1

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))

Answer 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

Answer 3

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:

• itertools.combinations
• any()
• unrolling (complex) list comprehensions

Answer 4

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 .

Answer 5

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

Answer 6

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