Subtracting consecutive elements in a list when the first element is great than the next - Python

Question

I'm making a program which completes the following:

When inputted a list a, it will subtract consecutive elements (starting from the beginning) if the results are non-negative numbers. For example, if

a=[3,2,1]

then consecutive numbers will be subtracted, so a=[1,1], and then a=[0]. Also, in the result, all numbers must be ascending (for example a 2,1 cannot be in the list). Another example:

a=[1, 10, 7, 3, 2]
[1, 3, 3, 2] #10-7 (10 and 7 get replaced with 3)
[1, 0, 2]    #3-3  (precedence goes to the left: 3-3 gets subtracted, not 3-2)
[1, 2]       #1-0

Here is my current code (where a is being generated randomly):

import random
a=[random.randint(1,10) for e in range(20)]
print(a)
loop=1
while True:
    try:
        #print(loop,a)
        subloop=0
        while subloop<loop:
            if a[loop-subloop]<=a[loop-1-subloop]:
                a[loop-1-subloop]=a[loop-1-subloop]-a.pop(loop-subloop)
                if loop!=1:
                    loop-=1
            subloop+=1
        if a[loop]<=a[loop-1]:
            a[loop-1]=a[loop-1]-a.pop(loop)
        else:
            loop+=1
    except IndexError:
        break
print(a)

This code feels kind of long/ineffiecient. Is there a better or more short/efficient way to do this?

Answer 1

Here's my take on it:

a = [1, 10, 7, 3, 2]
b = [3, 2, 1]


def index_helper(l):
    for i, x in enumerate(l[:-1]):
        if l[i] >= l[i+1]:
            return i


def reduce(l):
    i = index_helper(l)
    while i is not None:
        l[i:i + 2] = [l[i] - l[i + 1]]
        i = index_helper(l)
    return l


>>> reduce(a)
[1, 2]

>>> reduce(b)
[0]