CODEWITHSUNDEEP
python
Sahaj Bajaj
Sahaj Bajaj

Reputation: 1

Write a function sum13(nums)

Write a function sum13(nums) that takes a list of ints nums and returns the sum of the numbers in the list, returning 0 for an empty list. Except the number 13 is very unlucky, so it does not count and numbers that come immediately after a 13 also do not count.

Test    Result
print(sum13([1, 2, 2, 1, 13]))
6
print(sum13([13, 2, 2, 1]))
3
print(sum13([13, 13, 2, 13, 1, 5, 6, 13]))
11(EXPECTED)
13(GOT)

here is my program I don't know why the last test code is not running

def sum13(nums):
    """jdhdjkls"""
    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)

Upvotes: 0

Views: 225

Answers (4)

tomo_iris427
tomo_iris427

Reputation: 158

Your scripts can't correspond to 13 13 2 because they are made 0 0 2 in first step.

for and continue is effective for your goal.

def sum13(nums):
    sums=0
    after13=False
    for num in nums:
        if num == 13:
            after13=True
            continue
        if not after13:
            sums+=num
        after13=False
    return sums

Upvotes: 0

cdlane
cdlane

Reputation: 41905

We could try using recursion to solve this problem:

def sum13(numbers):
    if not numbers:  # base case
        return 0

    first, *rest = numbers

    if first == 13:
        first = 0

        if rest and rest[0] != 13:
            rest.pop(0)

    return first + sum13(rest)

print(sum13([1, 2, 2, 1, 13]))
print(sum13([13, 2, 2, 1]))
print(sum13([13, 13, 2, 13, 1, 5, 6, 13]))

Though not the style of answer you likely want, we can also coerce itertools.groupby to do the work for us:

from itertools import groupby

def sum13(numbers):
    previous13 = False

    def is_tainted(number):
        nonlocal previous13

        tainted = previous13
        previous13 = number == 13

        return True if tainted else previous13

    total = 0

    for tainted, subset in groupby(numbers, is_tainted):
        if not tainted:
            total += sum(subset)

    return total

Upvotes: 0

Shadowcoder
Shadowcoder

Reputation: 972

This will solve your problem:

def sum13(nums):
    """jdhdjkls"""
    if len(nums) == 0:
        return 0

    for i in range(0, len(nums)):
        if nums[i] == 13:
            nums[i] = 0
            # problem arises when there are 2 consecutive 13 this if statement will rectify it.
            if (i+1 < len(nums)) and (nums[i+1]!=13): 
                nums[i+1] = 0
    return sum(nums)

Upvotes: 0

Red
Red

Reputation: 27577

You can use the built-in zip() method to zip two lists skewed like:

[1, 2, 3, 4]
[1, 2, 3, 4]

to

[1, 2, 3, 4]
[2, 3, 4, None]

and check to see if any of the pairs have 13 in them.

Here is how:

def sum13(nums):
    s = 0 if nums[0] == 13 else nums[0]
    for p in zip(nums, nums[1:]):
        if all(n != 13 for n in p):
            s += p[1]
    return s

print(sum13([1, 2, 2, 1, 13]))
print(sum13([13, 2, 2, 1]))
print(sum13([13, 13, 2, 13, 1, 5, 6, 13]))

Output:

6
3
11

You can also unpack the pairs:

def sum13(nums):
    s = 0 if nums[0] == 13 else nums[0]
    for i, j in zip(nums, nums[1:]):
        if i != 13 != j:
            s += j
    return s

print(sum13([1, 2, 2, 1, 13]))
print(sum13([13, 2, 2, 1]))
print(sum13([13, 13, 2, 13, 1, 5, 6, 13]))

Upvotes: 1

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