CODEWITHSUNDEEP
python
tekemsi
tekemsi

Reputation: 3

Trying to sum numbers between numbers more efficiently

TLDR: Need to make code below more efficient.

I'm trying to solve a coding challenge that requires I sum all numbers between two numbers. My code shown below is passing the tests but is taking too long to run, causing me to fail due to a time out.

Is there any way to rewrite this to do the same thing in less time?

EDIT: solved, thank you so much for your solutions

def get_sum(a,b):
    res = 0
    if b>a:
        for i in range(a,b):
            res = sum(range(a,b+1))
            return res
    if a > b: 
        for i in range (b,a):
            res = sum(range(b,a+1))
        return res
    elif a == b:
        return a

Upvotes: 0

Views: 68

Answers (4)

ringo
ringo

Reputation: 1158

Yup. Using the formula for the triangular numbers:

def get_sum(a,b):
    if a > b:
        b, a = a, b
    sum_from_1_to_a = a * (a + 1) / 2
    sum_from_1_to_b = b * (b + 1) / 2
    return sum_from_1_to_b - sum_from_1_to_a

You can simplify this, obviously, but I left it this way so it is obvious how the formula is being applied.

Also, are you sure if a == b you should return a? Aren't there 0 numbers between them if they have the same value?

Upvotes: 0

Abdel-hilmi
Abdel-hilmi

Reputation: 73

One suggestion, when you find yourself trying to check for the order of parameters do something like this so you don't have to write the same code twice:

if a>b:
  a,b = b,a

# your code

Upvotes: 0

zooid
zooid

Reputation: 101

Maybe this?

def get_sum2(a, b):
    if a == b:
        return a
    return sum([n for n in range(a, b+1)] if b > a else [n for n in range(b, a+1)])

if __name__=="__main__":
    print(get_sum2(1, 3))

Upvotes: 0

ozdemir
ozdemir

Reputation: 389

a little math won't harm.

enter image description here

so this should work.

def get_sum(a,b):
    return (b-a+1)*(a+b)/2

Upvotes: 3

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