If I had the 1D list:

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

How would I separate the 1D list so that it would make a 2D list that has 1D elements of decreasing length, down to 1?

For example, the above list would become:

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

I am aware this only works with lists of certain lengths, but in this case I only need it to work for those ideal lengths.

CodePudding user response：

if no other solution is coming to your mind ,just start forming new lists from the end, it will be an O(n) solution.

CodePudding user response：

With a generator

Let's make a generator which will return the elements of a list as if they were in a flattened triangular matrix:

def tril(arr):
    '''Iterate over the list as a lower triangular matrix,
    e.g. if arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] then return one by one:
    [1]
    [2, 3]
    [4, 5, 6]
    [7, 8, 9, 10]
    '''
    N = len(arr)
    left, width = 0, 1
    while left < N:
        yield arr[left:(right:=left width)]
        left = right
        width  = 1

With [i for i in tril([*range(1,11)]) we get a list:

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

So to get the required answer we have to apply reversing three times - to a given list arr = [*range(1,11)], to each item from tril(reversed(arr)), and to the whole output list:

arr = [*range(1, 11)]
answer = [i[::-1] for i in tril(arr[::-1])][::-1]

Knowing the lenght of the first item

We can calculate the length of the first item from the equation:

n*(n 1) == 2*len(arr)

where arr is a given list if items, and n is the lenght of the first item of the required answer, e.g if arr = [*range(1, 11)] then n == 4. In other words, len(arr) is equal to the sum of first n natural numbers.

So we can obtain the answer this way:

arr = [*range(1, 11)]
n = round(((8*len(arr)   1)**0.5 - 1)/2)

# Usually we know the value of n
# so only the next 2 lines is needed in common case
item = iter(arr)
answer = [[next(item) for _ in range(i)] for i in range(n, 0, -1)]