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Your solution is well-implemented and correct. Here are some points for improvement:
Time Complexity Comment: The comment for time complexity is not accurate. The worst-case time complexity is actually exponential. A better way to describe it might be O(k * 2^t), where k is the average length of combinations, but it's complex. You could say it's exponential in the target value.
Space Complexity Comment: The space complexity is O(t) for the recursion stack, but note that the output storage (the result list) requires additional space, which is O(n * number_of_combinations). So the overall space complexity is higher. However, the recursion stack space is O(t), which is correct.
Sorting and Early Break: Sorting the candidates and breaking early when the candidate exceeds the target is an excellent optimization. This reduces unnecessary recursion.
Code Readability: The code is clean and easy to understand. Using a for-loop with index tracking is a standard approach for combination problems. The use of path.copy() ensures that we store a snapshot of the current path, which is correct.
Efficiency: The solution is efficient. However, note that the reference solution uses a different approach (choose/not choose) which might be less efficient in terms of space because it creates new lists at each step. Your approach uses backtracking with a single list that is modified in place, which is more space-efficient.
Minor Suggestion: Consider adding a docstring to the helper function to explain its parameters, as it improves readability.
Overall, great job! The solution is optimal and follows best practices.
The solution correctly implements the backtracking approach with a for-loop, which is efficient and avoids duplicate combinations by using the index.
The code is clean and readable, with good variable names and structure.
The student has included time and space complexity analysis, which shows understanding of the algorithm's efficiency.
Areas for Improvement:
Time Complexity Analysis: The student's time complexity of O(n^t) is a bit vague. A more precise analysis would note that the worst-case time complexity is exponential, specifically O(n^(target/min_candidate)), which is standard for backtracking in combination sum problems. However, the provided O(n^t) is acceptable as a high-level estimate.
Space Complexity: The student states O(t), which refers to the depth of the recursion stack (which is at most target/min_candidate). However, the space complexity should also account for the storage of the result, which can be significant. The reference solution notes O(n^2) for space due to deep copies, but in this solution, the backtracking uses a single path list that is modified in-place, so the space for the recursion stack is O(t), but the result storage is O(n * number_of_combinations). It's important to mention both.
Unnecessary Sorting: The solution sorts the candidates, which is not strictly required for correctness but can help in optimization (to break early when candidates exceed the target). This is a good optimization, but it adds O(n log n) time, which is acceptable given the constraints.
Deep Copies: The solution uses path.copy() when adding to the result, which is efficient because it only copies when a valid combination is found. This is better than creating deep copies at every recursive call (as in the reference solution), so the student's approach is more space-efficient.
Suggestions:
Consider adding a comment explaining why the sorting is done (to enable the break condition).
The space complexity comment could be expanded to note that the recursion stack depth is O(t) and the result storage is O(n * k) where k is the number of combinations.
Overall, the solution is well-implemented and efficient.
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