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from collections import deque
def move_zeroes_to_end_quadratic(xs):
"""
This solution is suboptimal. It runs in quadratic time, and it uses constant
space.
"""
i = 0
while i < len(xs) - 1:
if xs[i] == 0:
j = i + 1
while j < len(xs) and xs[j] == 0:
j += 1
if j >= len(xs):
break
xs[i], xs[j] = xs[j], xs[i]
i += 1
def move_zeroes_to_end_linear(xs):
"""
This solution is clever. It runs in linear time proportionate to the number
of elements in `xs`, and has linear space proportionate to the number of
consecutive zeroes in `xs`.
"""
q = deque()
for i in range(len(xs)):
if xs[i] == 0:
q.append(i)
else:
if q:
j = q.popleft()
xs[i], xs[j] = xs[j], xs[i]
q.append(i)
def move_zeroes_to_end_linear_constant_space(xs):
"""
This is the optimal solution. It runs in linear time and uses constant
space.
"""
i = 0
for j in range(len(xs)):
if xs[j] != 0:
xs[i], xs[j] = xs[j], xs[i]
i += 1
################################################################################
# Tests
################################################################################
xss = [
[1, 2, 0, 3, 4, 0, 0, 5, 0],
[0, 1, 2, 0, 3, 4],
[0, 0],
]
f = move_zeroes_to_end_linear_constant_space
for xs in xss:
print(xs)
f(xs)
print(xs)
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