我的问题是关于我描述的代码但它不起作用

Question

SUMMER OF '69: Return the sum of the numbers in the array, except ignore sections of numbers starting with a 6 and extending to the next 9 (every 6 will be followed by at least one 9). Return 0 for no numbers.¶

I have tried to use the pop method but it didn't work. I want to know why.

def summer_69(arr):
    num=(6,7,8,9)
    if num not in arr:
        return sum(arr)
    if num in arr:
        arr.pop(num)
        return sum(arr)

print(summer_69([4,5,6,7,8,9]))

I am getting the whole sum like in this one I am getting like 39.

Answer 1

Pop removes the item at a specified index. Max index in your array is 5 (4 has index 0). I recommend finding the index of 6 via

arr.index(6)

Not efficient, but you can then repeatedly pop that index until it becomes 9, (and popping once more if "extending to" means including the 9).

Answer 2

Instead of doing this i will suggest you to use a single loop as it will be more efficient. Use single loop and keep adding number to sum until 6 is encountered. As soon as 6 comes skip the numbers till a 9 occurs. Again start to add numbers to sum. Runs in O(n).

i=0

Sum =0

While i < length:

If a[i] == 6:

     While i < length and a[i] != 9:

         i += 1

Else:

    Sum += a[i]

i += 1

return Sum