Return remaining items in list, if last n elements in list greater than a number

Question

Given a list of numbers:

ar =[1,2,3,4,5,6,7,8,9,10]

I need to find the remaining part of the list if the last n numbers summed are greater than a given number. For example if the given number was 25, the returned\\"remainder" list would be [1,2,3,4,5,6,7] , since that the last 3 elements add up to 27.

I have some code, but I get the overbearing suspicion that it is quite inefficient, failing that, overly verbose given the simple nature of the task.

def kp_find(ar, level):

    kp1=[]    
    for k in range(len(ar)): #this is required as in reality there is some
        kp1.append(k)        #work to be done within this loop 

    kp_fin = sorted(kp1,reverse = True)

    kp2=[]
    for k in kp_fin:
        kp2.append(k)
        sm_it = sum([i for i in kp2])
        if sm_it>level:
            place = [i for i, j in enumerate(kp_fin) if j == k]
            break
        else:
            place = []

    if place:
        return kp_fin[place[0]:len(kp1)]

print kp_find(ar,24)

As you see it's quite a lot of code for something so insignificant; any help welcome.

Answer 1

It does seem like a lot of code for a simple task. You could just pop off elements from the right end of the list until the condition is met. Something like this:

>>> def kp_find(ar, level):
...     result = ar[:]
...     sum_ = 0
...     while sum_ < level:
...         sum_ += result.pop()
...     return result
... 
>>> kp_find([1,2,3,4,5,6,7,8,9,10], 25)
[1, 2, 3, 4, 5, 6, 7]