python - Split array to approximately equal chunks

Question

How split array into two chunks, when sum of every chunk is approximately equal?

>>> foo([10, 1, 1, 1])
[[10], [1, 1, 1]]
>>> foo([2, 5, 9, 5, 1, 1])
[[2, 5], [9, 5, 1, 1]]
>>> foo([9, 5, 5, 8, 2, 2, 18, 8, 3, 9, 4])
[[9, 5, 5, 8, 2, 2], [18, 8, 3, 9, 4]]
>>> foo([17, 15, 2, 18, 7, 20, 3, 20, 12, 7])
[[17, 15, 2, 18, 7], [20, 3, 20, 12, 7]]
>>> foo([19, 8, 9, 1, 14, 1, 16, 4, 15, 5])
[[19, 8, 9, 1], [14, 1, 16, 4, 15, 5]]

Answer 1

You can create your slicees with loop over your list then choose the proper pairs with min function with a proper key :

>>> def find_min(l):
...     return min(((l[:i],l[i:]) for i in range(len(l))),key=lambda x:abs((sum(x[0])-sum(x[1]))))

Demo :

>>> l=[10, 1, 1, 1]
>>> find_min(l)
([10], [1, 1, 1])
>>> l=[9, 5, 5, 8, 2, 2, 18, 8, 3, 9, 4]
>>> find_min(l)
([9, 5, 5, 8, 2, 2], [18, 8, 3, 9, 4])
>>> l=[19, 8, 9, 1, 14, 1, 16, 4, 15, 5]
>>> find_min(l)
([19, 8, 9, 1, 14], [1, 16, 4, 15, 5])