Sum of numbers in a Python list except 13 and numbers after it

Question

Return the sum of the numbers in the array, returning 0 for an empty array. Except the number 13 is very unlucky, so it does not count and numbers that come immediately after a 13 also do not count.

def sum13(nums):
  if nums == []:
    return 0
  sum=0
  for i in range(len(nums)):
    if nums[i] == 13:
      break
  else:
    sum = sum+nums[i]
  return sum

The above code is failing only at:

sum13([1, 2, 13, 2, 1, 13]) → 4 while my answer is 3 (how 4?)
sum13([13, 1, 2, 13, 2, 1, 13]) → 3 while my answer is 0 (again, how is 3 correct if there are no numbers before 13?)
Please explain the logic and what I’ve done wrong.

Asked By: Bharath Kalyan

||

Source

Answer 1

Numbers "immediately after a 13" means that any number directly following a 13 in the array is ignored. For example, in [1, 2, 13, 2, 1, 13], the 2 after the 13 is ignored, but the 1 after that 2 counts.

Answered By: rikusv

Answer 2

The numbers that come immediately after 13, not ALL the numbers after 13.

def sum13(nums):
  if nums == []:
    return 0
  sum=0
  i = 0
  while i <= len(nums): #change this to a while loop so we can skip numbers
    if nums[i] == 13:
      i+=1 # if it's 13 we skip an extra number
    else:
      sum = sum+nums[i]
    i+=1 # this always increments the counter, but then on 13 we are now incrementing twice

  return sum


print(sum13([1, 2, 13, 2, 1, 13]))

Answered By: Rashid 'Lee' Ibrahim

Answer 3

If you´re surfing through stacks trying to find some elegant solution, here´s mine:

def sum13(nums):

  if nums == 0:
    return 0
  soma = 0
  i = 0
  
  while i < len(nums):  
    if nums[i] == 13:
      i += 2
      continue
    else: 
      soma += nums[i]
      i += 1
  return soma

Answered By: Jean Paul Dosher

Answer 4

def damned_13(nums):
      for i in nums:
            if i == 13:
                  return sum(nums[:nums.index(13)])
      return sum(nums)
print(damned_13([1, 2, 2, 1, 13, 55]))

or

def damned_13(nums):
      if nums.count(13) == 1:
            return sum(nums[:nums.index(13)])
      return sum(nums)
print(damned_13([1, 2, 2, 1, 13, 55]))

Answered By: ogk

Answer 5

a=['1','2','13','2','1','13']
b=a.index('13')
a=a[:b]
a=list(map(int,a))
print(sum(a))

Answered By: İlaha Algayeva

Sum of numbers in a Python list except 13 and numbers after it

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