How to overcome maximum recursion depth exceeded error here?

Question

The problem statement is as follows:

Return the sum of the numbers in the array, except ignore sections of numbers starting with a 6 and extending to the next 7 (every 6 will be followed by at least one 7). Return 0 for no numbers.

sum67([1, 2, 2]) → 5
sum67([1, 2, 2, 6, 99, 99, 7]) → 5
sum67([1, 1, 6, 7, 2]) → 4

I'm aware of the same problem being posted here on stack overflow but I don't want to request a new solution, rather I want to know what might be the problems with my own recursive solution to the problem.

My attempt:

def sum67(nums):
  def rem67(nums):
    if 6 not in nums:
      return nums
    index6 = nums.index(6)
    index7 = nums.index(7)
    nums = nums[:index6] + nums[(index7 + 1):]
    return rem67(nums)
  return sum(rem67(nums))

The server keeps throwing maximum recursion depth exceeded error which I'm unable to rectify on the server. Any leads will be appreciated.

Your rem67 will break if the first 7 comes before the first 6. — 0x5453
– 0x5453, Commented Jul 2, 2021 at 12:56

Sefan · Accepted Answer · 2021-07-02 13:09:51Z

2

The recursion limit is 1000 by default. You could use a while true instead, like:

def sum67(nums):
    while True:
        if 6 not in nums:
            break
        index6 = nums.index(6)
        index7 = nums.index(7)
        nums = nums[:index6] + nums[(index7 + 1):]
    return sum(nums)

Like 0x5453 commented, it breaks if you have a 7 before a 6. So you probably want to add something like this:

if index6 > index7:
    return -1

edited Jul 2, 2021 at 13:09

answered Jul 2, 2021 at 13:01

Sefan

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1 Comment

iamakhilverma Over a year ago

while True doesn't work as it is like you've shown but I figured out a way around it. Thank you

Pippet39 · Accepted Answer · 2021-07-02 13:51:01Z

2

If a 7 occurs in the array before its 6 counterpart, then on one iteration the new array formed in rem67() will be composed of a section with a 7 but no 6, followed by a section with a 6. This sets up the same condition in the new array, which will result in infinite recursion.

If it is allowed that two consecutive 7's may occur, or for a 7 to occur first, then it would be better to do the following:

index7 = nums.index(7, index6)

This makes index7 reference the first 7 that follows the 6, thus ignoring any 7's that may occur before the 6.

Otherwise returning an error code as suggested by Sefan is good.

Hope this fixes things :)

edited Jul 2, 2021 at 13:51

answered Jul 2, 2021 at 13:07

Pippet39

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Comments

Alexandre Simões · Accepted Answer · 2021-07-02 13:07:28Z

0

This isn't a safe mode:

import sys

my_limit = sys.getrecursionlimit()
sys.setrecursionlimit(my_limit + 200)
print(sys.getrecursionlimit())

answered Jul 2, 2021 at 13:07

Alexandre Simões

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iamakhilverma · Accepted Answer · 2021-07-02 14:48:49Z

0

As per @0x5453 's comment, I modified the index7 which now looks for only those 7 which come after we encounter 6.

Here's the modified code which works correctly:

def sum67(nums):
  def rem67(nums):
    if 6 not in nums:
      return nums
    index6 = nums.index(6)
    index7 = nums[index6:].index(7) + index6
    nums = nums[:index6] + nums[(index7 + 1):]
    return rem67(nums)
  return sum(rem67(nums))

answered Jul 2, 2021 at 14:48

iamakhilverma

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