looking for a more elegant solution to this

python

Solution

If we can just remove the elements that we don't want, then we can use a simple sum. Here is an example:

def sum67(nums):
    nums=nums[:]
    while 6 in nums:
        i=nums.index(6)
        j=nums.index(7,i)
        del nums[i:j+1]
    return sum(nums)

First, we use `nums=nums[:]` to make a copy. The caller probably isn't expecting `nums` to change.

`nums.index(6)` finds the index of the first element that has a value of 6. `nums.index(7,i)` finds the index of the first element that has a value of 7 after the index `i`. `del nums[i:j+1]` then deletes the elements in the range from `i` to `j`, including the element at `j`.

Problem

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 ``` ``` def sum67(nums): dontadd = 0 sum = 0 for i in range(0, len(nums)): if dontadd == 0: if nums[i] == 6: dontadd = 1 else: sum += nums[i] else: if nums[i] == 7: dontadd = 0 else: pass# nothing happens. It is useful as a placeholder when a statement is required syntactically return sum ``` Looking for a more elegant solution to this problem from codingbat. This answer doesn't seem as intuitive as it could be

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