Permutations Leetcode Solution

class Solution {
public:
    #define DPSolver ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);

void permutations(const vector< int> &nums, vector< vector< int>> &res, vector< bool> &taken, vector< int> per, const int sz, int lvl)
{
    if (lvl == sz)
    {
        res.push_back(per);
        return;
    }
    for (int i = 0; i < sz; i++)
    {
        if (!taken[i])
        {
            taken[i] = true;
            per.push_back(nums[i]);
            permutations(nums, res, taken, per, sz, lvl + 1);
            per.pop_back();
            taken[i] = false;
        }
    }
}

vector< vector< int>> permute(vector< int> &nums)
{
    DPSolver;
    vector< vector< int>> res;
    vector< bool> taken(nums.size());
    vector< int> per;
    permutations(nums, res, taken, per, nums.size(), 0);
    return res;
}

    vector< vector< int>> permuteUsingBuiltInFunction(vector< int>& nums) {
        // to get all the possible permutations.
        sort(nums.begin(), nums.end());
        
        // creating a place in the memory to store the results in. 
        vector< vector< int>> permus; 

        // generating all possible combinations. 
        do {
            permus.push_back(nums); 
        } while (next_permutation(nums.begin(), nums.end()));

        // returning the result. 
        return permus; 
    }

};

public List< List< Integer>> subsets(int[] nums) {
    List< List< Integer>> list = new ArrayList<>();
    Arrays.sort(nums);
    backtrack(list, new ArrayList<>(), nums, 0);
    return list;
}

private void backtrack(List< List< Integer>> list , List< Integer> tempList, int [] nums, int start){
    list.add(new ArrayList<>(tempList));
    for(int i = start; i < nums.length; i++){
        tempList.add(nums[i]);
        backtrack(list, tempList, nums, i + 1);
        tempList.remove(tempList.size() - 1);
    }
}

class Solution:
    def permute(self, nums: List[int]) -> List[List[int]]:
        ans = []

        def backtrack(path):
            if len(path) == len(nums):
                ans.append(path[:])
                return 

            for num in nums:
                if num not in path:
                    path.append(num)
                    backtrack(path)
                    path.pop()

        backtrack([])

        return ans