CODEWITHSUNDEEP
pythonrecursion
Richard Coterie
Richard Coterie

Reputation: 17

How to get the sum of a list of numbers excluding integers that are divisible by 3 and 7 with recursion?

I am trying to find the summation of integer in list with elements that are divisible by 3 or 7 excluded

def SumSkip37(numList,sum = 0):
    if numList:
        i = numList.pop()
        if i % 3 == 0 or i % 7 == 0:
            return sum
        else:
            sum += i
            return SumSkip37(numList, sum=sum)



numList = [1, 3, 5, 7, 9]
print(f'The result is {SumSkip37(numList)}.')

Pls help me figure out

Upvotes: 0

Views: 620

Answers (4)

user2390182
user2390182

Reputation: 73480

On relatively clean way:

def SumSkip37(numList):
    if not numList:
        return 0
    head, *tail = numList
    return head * bool(head % 3 and head % 7) + SumSkip37(tail)

Upvotes: 0

mozway
mozway

Reputation: 261934

You can update your code to:

def SumSkip37(numList, my_sum = 0): # avoid using sum
    if numList:
        i = numList.pop()
        if i % 3 == 0 or i % 7 == 0:
            return SumSkip37(numList, my_sum=my_sum) # pass the unchanged sum
        else:
            return SumSkip37(numList, my_sum=my_sum+i) # pass the sum + i
    else:
        return my_sum

NB. I tried to stay as close as possible to the original, but you can simplify greatly!

To avoid mutating the input:

def SumSkip37(numList, my_sum = 0):
    if numList:
        if numList[0] % 3 != 0 and numList[0] % 7 != 0:
            my_sum += numList[0]
        return SumSkip37(numList[1:], my_sum=my_sum)
    return my_sum
print(f'The result is {SumSkip37(numList)}.')

Better approach than the alternative above suggested by @Stef to run in linear time:

def SumSkip37(numList, my_sum=0, idx=0):
    if idx >= len(numList):
        return my_sum
    elif numList[idx] % 3 != 0 and numList[idx] % 7 != 0:
        return SumSkip37(numList, my_sum + numList[idx], idx + 1)
    return SumSkip37(numList, my_sum, idx + 1)

Also, recursion is overkill here, better use a short generator expression:

sum(i for i in numList if i%7!=0 and i%3!=0)

Upvotes: 2

Turhan Can Kargin
Turhan Can Kargin

Reputation: 26

Without Recursive function:

def SumSkip37(list):
    sum = 0
    for i in list:
        if i % 3 == 0 or i % 7 == 0:
            continue
        else:
            sum += i
    return sum

list = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
print(SumSkip37(list))

With Recursive Function:

def SumSkip37(lst):
    if len(lst) == 0:
        return 0
    else:
        if lst[0] % 3 == 0 or lst[0] % 7 == 0:
            return SumSkip37(lst[1:])
        else:
            return lst[0] + SumSkip37(lst[1:])

print(SumSkip37([1, 2, 3, 4, 5, 6, 7, 8, 9, 10]))

Upvotes: 0

LTJ
LTJ

Reputation: 1255

You are on the right track. You just weren't iterating through the whole list:

def SumSkip37(numList, total=0):
    if numList:
        i = numList.pop()
        if i % 3 != 0 and i % 7 != 0:
            total += i
        return SumSkip37(numList, total=total)
    return total

I flipped the comparison to remove the unnecessary else branch. It doesn't matter if the current number is divisible with 3 or 7 or not, you always want to continue the recursion until the list runs out.

Upvotes: 0

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