'Sum of numbers in array, not counting 13 and number directly after it (CodingBat puzzle)

The 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.

My Code:

def sum13(nums):
  l = len(nums)
  tot = 0

  if l==0:
    return 0

  for x in range(l):
    if nums[x]!=13:
      if nums[x-1]!=13:
        tot+=nums[x]

  return tot

Where It's Failing:

sum13([1, 2, 2, 1, 13]) should → 6, but my code is outputting 5
sum13([1, 2, 13, 2, 1, 13]) should → 4, but my code is outputting 3

python list

Solution 1:^[1]

Your problem is when x is zero. x - 1 will be -1 so it will get the last element of your list (13). To fix it, don't test x - 1 if x is zero:

if x == 0 or nums[x-1] != 13:

In Python, when you pass a negative index to a list it accesses its elements backwards, so:

>>> x = [1,2,3,4,5]
>>> x[-1]
5
>>> x[-2]
4

Solution 2:^[2]

enumerate() can give the index, then it can be done using a generator expression in sum(). The extra 0 added to the end of the list makes it work for nums[i-1] when i = 0.

def sum13(nums):
   nums += [0]
   return sum(n for i, n in enumerate(nums) if n != 13 and nums[i-1] != 13)

Solution 3:^[3]

Here is the solution:

def sum13(nums):

    if len(nums) == 0:
        return 0

    for i in range(0, len(nums)):
        if nums[i] == 13:
            nums[i] = 0
            if i+1 < len(nums): 
                nums[i+1] = 0
    return sum(nums)

Solution 4:^[4]

You could avoid the bug by using next() function instead of error-prone indexes:

def exclude13(iterable):
    it = iter(iterable)
    for x in it:
        if x == 13: # don't yield 13
           next(it) # skip number immediately after 13
        else:
           yield x

print(sum(exclude13([1, 2, 2, 1, 13]))) # -> 6
print(sum(exclude13([1, 2, 13, 2, 1, 13]))) # -> 4

Solution 5:^[5]

def sum13(nums):
    count = 0
    while count < len(nums):
        if nums[count] == 13:
            del nums[count:count+2]
            continue
        count += 1
    return sum(nums)

Solution 6:^[6]

def sum13(nums):
    count = 0
    sum = 0
    for i in range(len(nums)):
         if nums[i] == 13 or sum == 1 :
            nums[i]= 0
            sum = sum + 1
         count = count + nums[i]
         if sum  ==  2:
            sum  = 0
    if nums == []:
          return 0
    return coun

Solution 7:^[7]

def sum13(nums):
  if len(nums)==0:
    return 0
  sum=0
  for i in range(1,len(nums)):
    if nums[i]==13 or nums[i-1]==13:
      sum = sum
    else:
      sum += nums[i]
  if nums[0]==13:
    return sum
  else:
    return sum + nums[0]

Start the loop form 1 and add back the first number (if it's not 13) at the end.

Solution 8:^[8]

Make a copy of the list so that the original doesn't change, remove instances of the number 13 and the number immediately following, then sum the resulting list:

def sum13(nums):
    nums = nums[:]
    while 13 in nums:
        i = nums.index(13)
        del nums[i:i+2]
    return sum(nums)

This is similar to the top answer for this problem: looking for a more elegant solution to this

Solution 9:^[9]

Another way to look at it:

def sum13(nums):
  sub = 0
  for i in range(len(nums)):
    if nums[i] == 13 :
      sub+=nums[i]
      if i+1<len(nums) and nums[i+1]!=13:
        sub+=nums[i+1]

  return sum(nums)-sub

Solution 10:^[10]

    def sum13(nums):
        sum = 0
        len_n = len(nums)
        k = 0
        while k < len_n:
            if nums[k] == 13:
                k += 2
            else:
                sum += nums[k]
                k += 1
        return sum

Solution 11:^[11]

sum of list except when item=13 will not be counted and item+1 also be not counted

def sum13(nums):
  if len(nums)==0:
    return 0
  if nums[-1]==13:
    nums[-1]=0
  for i in range(len(nums)-1):
    if nums[i]==13:
      nums[i]=0
      nums[i+1]=0
  return sum(nums)

Solution 12:^[12]

def sum13(nums):
  result = 0
  i = 0

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

  return result

Solution 13:^[13]

You're excluding numbers that come immediately before a 13 when you should be excluding numbers that come immediately after.

Solution 14:^[14]

I would suggest to use a recursive solution. Note that while index of a list can be out of range the slicing gives an empty list and not an error, e.g. if a = [1,1,2] then a[3] would give an error while a[3:] gives the empty list [].

def sum13(nums):
  if len(nums) == 0:
    return 0
  sum = sum13(nums[1:])
  if nums[0] != 13:
    return sum + nums[0]
  if nums[0] == 13:
    return sum13(nums[2:])

Solution 15:^[15]

I solved the problem in the following fashion. It is brute force, but it does work:

def sum13(nums):
    count = 0
    if len(nums) == 0: 
        return 0
    for ans in range(len(nums)):
        if nums[ans] == 13 and ans < len(nums)-1: 
            nums[ans] = 0
            nums[ans+1] = 0
        elif nums[ans] == 13 and ans == len(nums)-1:
            nums[ans] = 0 
        else: 
            count += nums[ans]

    return count

Solution 16:^[16]

def sum13(nums):
  nums2 = []
  if 13 in nums:
    for i in range(len(nums)-1): # the -1 is needed to avoid out of range.
      if nums[i] != 13: nums2.append(nums[i])
      else:
        nums[i] = 0
        nums[i+1] = 0
    return sum(nums2)
  return sum(nums)

I created a separate list and appended the figures i wanted into that list. i did this through a for loops by going through every element and appending that figure if it did not equal 13.

when it did eventually equal the number 13 i would then change that number by setting it to zero (nums[i] = 0). Then I also need to set the next number to it at Zero by indexing (nums[i+1] = 0). so they are then discounted from the sum of the list.

this is not a perfect solution but it is a working one.

Solution 17:^[17]

This is an approach that i have taken,this works as well

nums = [1, 2, 2, 1, 13,2,1,2,3,13,4,5]
counter = 0
temp = []
for i in range(len(nums)):
    if nums[i] == 13:
        if (i+1) != len(nums) and nums[i+1] != 13:
            temp.append(nums[i+1])
            continue
        elif (i+1) == len(nums):
            continue
    else:
        counter = counter + nums[i]
temp_total = sum(temp)
count_total = counter - temp_total
print(count_total)

Solution 18:^[18]

When x=0 in nums[x-1] it conflicts, because it results as nums[0-1] if it is 13 then it may not add so check nums[0] at last.

def sum13(nums):
  co=0
  if len(nums)<=0:
    return co
  for a in range(1,len(nums)):
    if nums[a]==13:
      continue
    elif nums[a-1]==13:
      continue
    else:
      co=co+nums[a]      
  if nums[0]!=13:
    co+=nums[0]
  return co

Solution 19:^[19]

This is my solution for this exercise!!!!

def sum13(nums):
    total = 0
    for i in range(len(nums)):
        if (nums[i] == 13 or nums[i-1] == 13 and i != 0):
            total = total
        else:
            total += nums[i]
    return total

Solution 20:^[20]

def sum13(nums):
    sum=0
    ind13=[]
    for num in nums:
        if num==13:
            num=0
            ind13=nums.index(13)
            ind13next=ind13+1
            nums[ind13]=0
            if nums[ind13] !=nums[-1]:
                nums[ind13next]=0
                sum=sum+num
        else:
            sum=sum+num
    return sum

Solution 21:^[21]

def sum13(nums):

  summed=0
  flag=True
  index_13=0

  for index, num in enumerate(nums):

    if num==13:
      index_13=index
      flag=False

    if flag==True:
      summed=summed+num

    if index_13+1<= len(nums)-1 and num==nums[index_13+1]:
       flag=True

  return summed

Solution 22:^[22]

def sum13(nums):
    li=[]
    if nums[-1]==13:
        li.append(13)
    for i in range(len(nums)-1):
        if nums[i]==13:    
            li.append(nums[i])
            li.append(nums[i+1])
    n=sum(nums)
    return (n-sum(li))

Solution 23:^[23]

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):
       nums += [0]
       add=sum(j for i, j in enumerate(nums) if j != 13 and nums[i-1] != 13)
       return add

Solution 24:^[24]

def sum13(nums):
  sum=0
  count=0
  for i in range(len(nums)):
    if nums[i]!=13:
        count=count-1
        if count<=0:
            sum=sum+nums[i]
    else:
        count=2
  return sum

Solution 25:^[25]

The way I think about the problem lends me to loop over the list - except for the last element. Then if a value is 13, the next element becomes 0. From there - I reload the list - and replace ALL 13 with 0. In the end I simply sum the values of the list.

def sum13(nums):
for i,x in enumerate(range(len(nums)-1)):
    if (nums[x]==13):
        nums[i+1]=0
        nums=nums
for i,x in enumerate(nums):
    if x==13: #if the element is equal to 13
        nums[i]=0
return sum(nums)

Solution 26:^[26]

This should work as well. Basically it's a double continue statement. You assign pairs of index and value for each iteration with enumerate(). If a number is == 13 --> continue, the next iterated item should qualify for the elif statement, which also continues for items preceded by a 13. Elif runs only for items at indices >= 1 (so there is no issue with going out of range at [0]).

def sum13(nums):
  if nums ==[]:
    return 0
  sum = 0
  for index, val in enumerate(nums):
    if nums[index] == 13:
      continue
    elif nums[index-1] == 13 and index > 0:
      continue
    else:
      sum += val
  return sum

Solution 27:^[27]

def sum13(nums):

  total = sum(nums)

  for i in range (len(nums)):
    if nums[i]==13:
      total = total - nums[i]
      if (i+1) < len(nums) and nums[i+1]!=13: #Using !=13 because 13 will already
        total = total-nums[i+1]               #be detected from previous if

  return total

Solution 28:^[28]

def sum13(nums):
  if len(nums) == 0:
    return 0
  total = 0
  for i in range(len(nums)):
    if nums[i] == 13 and  (len(nums)-1)>i:
      nums[i+1]=0
      total = total
    elif nums[i]==13:
      total = total
    else:
      total = total +nums[i]
  return total

Solution 29:^[29]

here another solution :

def sum13(nums):
  for i in range (len(nums)) :
    if nums[i] == 13 :
      nums[i] = 0 
      if i+1 < len(nums) and nums[i+1] != 13 :
        nums[i+1] = 0
        
  return sum(nums)

Solution 30:^[30]

This is how I have tried:)

sum = 0
for i in range(len(a)):
    if a[i] == 13:
        if i+1 < len(a):
            a[i], a[i+1] = 0, 0
        else:
            a[i] = 0
    else:
        sum += a[i]
return sum

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution	Source
Solution 1	mgibsonbr
Solution 2	dansalmo
Solution 3	Kapil Marwaha
Solution 4	jfs
Solution 5	Prasath
Solution 6	Taryn
Solution 7
Solution 8	Stuart Jennings
Solution 9	designDemon
Solution 10	Muldec
Solution 11	gehbiszumeis
Solution 12	Hong Z
Solution 13	Alex Hammel
Solution 14
Solution 15	rayryeng
Solution 16	Stefan T
Solution 17
Solution 18	vezunchik
Solution 19	ThachLeQuang
Solution 20	4b0
Solution 21
Solution 22	Megha Shyam
Solution 23	YashBora
Solution 24	ManojDS
Solution 25
Solution 26	pantomassi
Solution 27
Solution 28	sahil vamja
Solution 29	Dhia Ben Aicha
Solution 30	Pema Namgay

'Sum of numbers in array, not counting 13 and number directly after it (CodingBat puzzle)

Solution 1:[1]

Solution 2:[2]

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Solution 10:[10]

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Solution 19:[19]

Solution 20:[20]

Solution 21:[21]

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Solution 23:[23]

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Solution 27:[27]

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Solution 30:[30]