To solve this problem, we need to:
Let’s start with the Python function signature and then write the code step by step:
def alternating_groups(arr):
# Your logic here
pass
# Example to test:
example_arr = [sequence of numbers]
print(alternating_groups(example_arr))
We will determine the time complexity after implementing the function by analyzing each part of the code.
Now, let’s flesh out this plan:
def alternating_groups(arr):
result = []
group = []
add_group = True # Start by adding to the group
for num in arr:
if add_group:
group.append(num)
else:
result.append(num)
add_group = not add_group # Flip between adding to group and result
alternating_result = []
while group or result:
if group:
alternating_result.append(group.pop(0))
if result:
alternating_result.append(result.pop(0))
return alternating_result
# Example to test:
example_arr = [1, 2, 3, 4, 5, 6, 7, 8, 9]
print(alternating_groups(example_arr))
# Test additional edge cases to ensure robustness
print(alternating_groups([])) # Should return []
print(alternating_groups([1])) # Should return [1]
print(alternating_groups([1, 2])) # Should return [1, 2]
group
and result
lists runs in O(n) time.group
and result
into alternating_result
also runs in O(n) time.Hence, the overall time complexity is O(n), where n is the length of the input array.
group
and result
, each could store up to n elements.Thus, the space complexity is O(n).
By following this approach, we’ve created a clear and efficient solution for the problem.
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