Given a python list that contains nested lists of arbitrary levels of nesting, the goal is to return a completely flattened list i.e for the sample input [1, [2], [[[3]]], 1]
, the output should be [1, 2, 3, 1]
.
My solution:
def flatten(lst):
stack = [[lst, 0]]
result = []
while stack:
current_lst, start_index = stack[-1]
for i in range(start_index, len(current_lst)):
if isinstance(current_lst[i], list):
# Update the start_index of current list
# to the next element after the nested list
stack[-1][1] = i + 1
# Add nested list to stack
# and initialize its start_index to 0
stack.append([current_lst[i], 0])
# Pause current_lst traversal
break
# non list item
# add item to result
result.append(current_lst[i])
else:
# no nested list
# remove current list from stack
stack.pop()
return result
I would like to know the time and space complexity (auxiliary space) of my solution if correct.
What I think
Time Complexity:
I believe the solution has a time complexity of O(m + n) where m is the total number of nested lists at all levels and n is the total number of atomic elements at all levels (non list elements)
Space Complexity:
I believe the space complexity is O(d), where d is the depth of the most nested list. This is because the stack tracks the current state of traversal, and its size is proportional to the nesting depth
Is the solution correct?
Is the time and space analysis correct?
Yes, the solution is correct.
Yes, the time and space analysis are correct ... if you don't count the space used by result
as auxiliary space, which is reasonable. Although note that result
overallocates/reallocates, which you could regard as taking O(n) auxiliary space. You could optimize that by doing two passes over the whole input, one to count the atomic elements, then create result = [None] * n
, and then another pass to fill it.
Btw it's better to use iterators instead of your own list+index pairs (Attempt This Online!):
def flatten(lst):
stack = [iter(lst)]
result = []
while stack:
for item in stack[-1]:
if isinstance(item, list):
stack.append(iter(item))
break
result.append(item)
else:
stack.pop()
return result
Or with the mentioned space optimization (Attempt This Online!):
def flatten(lst):
def atoms():
stack = [iter(lst)]
while stack:
for item in stack[-1]:
if isinstance(item, list):
stack.append(iter(item))
break
yield item
else:
stack.pop()
n = sum(1 for _ in atoms())
result = [None] * n
i = 0
for result[i] in atoms():
i += 1
return result