We build permutations by choosing each unused element at every position. We use a used boolean array to track which elements have already been placed.

1. If the current permutation length equals n, save it.
2. Iterate through all elements in nums.
3. Skip elements already in the current permutation (tracked by used array).
4. Add the element, mark as used, recurse.
5. Backtrack: remove the element and unmark.

• Time Complexity: O(N! * N).
• Space Complexity: O(N).

def permute(nums):
    result = []
    
    def backtrack(current):
        if len(current) == len(nums):
            result.append(current[:])
            return
        for num in nums:
            if num not in current:
                current.append(num)
                backtrack(current)
                current.pop()
    
    backtrack([])
    return result