Backtracking 2 completed

[sujaygijre]

No description provided.

[super30admin]

Your solution for the subsets problem is correct and efficient. Well done! Here are a few points to consider for improvement:

1. The base condition in the backtrack function (checking if index == nums.size()) is unnecessary because the for loop condition (i < nums.size()) will naturally prevent further recursion when index reaches the end. You can remove that base condition without affecting the correctness. This will simplify the code slightly.

2. Currently, you are passing the temp vector by value. This means that at each recursive call, a new copy of the vector is made. While this is acceptable for small n (as n <= 10), it can be optimized by passing by reference and then using push and pop to backtrack. This would avoid the copying overhead and make the solution more efficient in terms of both time and space. For example:

   void backtrack(vector<int>& temp, vector<int>& nums, int index) {
        answer.push_back(temp);
        for (int i = index; i < nums.size(); i++) {
            temp.push_back(nums[i]);
            backtrack(temp, nums, i+1);
            temp.pop_back();
        }
   }

   Then call it with an empty temp initially. This change would reduce the time and space complexity of the recursive calls.

3. The variable name answer is a global variable. It is better to avoid global variables when possible. You could instead pass the result vector by reference or make it a member variable and clear it appropriately. However, in this context, since it's a class member, it is acceptable.

4. In the context of the problem, your solution is correct and efficient. The reference solution uses a different approach (without a for loop) but both are valid.

Overall, your solution is good. With the minor optimization of passing temp by reference and using push/pop, you can make it more efficient.