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In the following backtracking code snippet for subsets, which line causes all subsets in the result to be identical?

medium🐛 Buggy Code Q7 of Q15
Bit Manipulation - Subsets Using Bitmask
In the following backtracking code snippet for subsets, which line causes all subsets in the result to be identical?
def subsets(nums):
    res = []
    def backtrack(i, path):
        res.append(path)
        if i == len(nums):
            return
        backtrack(i + 1, path + [nums[i]])
        backtrack(i + 1, path)
    backtrack(0, [])
    return res
Ares.append(path)
BNo bug; code works correctly
Cbacktrack(i + 1, path)
Dbacktrack(i + 1, path + [nums[i]])
Step-by-Step Solution
Solution:
  1. Step 1: Analyze path usage

    Each recursive call creates a new list with path + [nums[i]] or passes path unchanged.
  2. Step 2: Check append

    Appending path (which is a new list each time) to res is safe; no shared references.
  3. Step 3: Confirm no mutation after append

    Since path is never mutated after append, subsets remain distinct.
  4. Final Answer:

    Option B -> Option B
  5. Quick Check:

    Using path + [nums[i]] creates new lists, avoiding shared references [OK]
Quick Trick: Using new lists avoids shared references [OK]
Common Mistakes:
MISTAKES
  • Assuming path is mutated in place
  • Confusing path + [nums[i]] with in-place append
  • Thinking append line causes bug without mutation
Trap Explanation:
PITFALL
  • Misunderstanding list immutability and copying leads to false bug suspicion.
Interviewer Note:
CONTEXT
  • Tests ability to identify subtle bugs related to list references in recursion.
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