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Update q2.py #1

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11 changes: 10 additions & 1 deletion q2.py
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Original file line number Diff line number Diff line change
Expand Up @@ -65,7 +65,16 @@ def compressed_stack_length(lst):
0
"""
# TODO: Implement your solution here
pass
count = 0
for i in range(len(lst)):
if lst[i] == lst[i-1]:
count = count - 1
else:
count = count + 1
if count < 0:
return 0
else:
return count
Comment on lines +68 to +77
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@gemini-code-assist gemini-code-assist bot Jan 19, 2026

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critical

The current implementation has a critical logic flaw and does not correctly model the stack-like behavior described in the problem.

  1. Incorrect Comparison: The code compares lst[i] with lst[i-1], which is the previous element in the original input list. The requirement is to compare with the "most recent active number", which is the top of the conceptual stack.
  2. Edge Case i=0: When i=0, lst[i-1] accesses the last element of the list (lst[-1]), which is incorrect. The first element should always be added to the stack as there's nothing to cancel it with.

A simple counter is insufficient to solve this problem because you need to know what the last active number was. A list used as a stack is the ideal data structure for this.

Here is a corrected implementation that uses a stack to accurately track the active numbers:

    stack = []
    for num in lst:
        if stack and stack[-1] == num:
            stack.pop()
        else:
            stack.append(num)
    return len(stack)



if __name__ == "__main__":
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