# Solving Leetcode Interviews in Seconds with AI: Permutations


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "46" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Given an array nums of distinct integers, return all the possible permutations. You can return the answer in any order.   Example 1: Input: nums = [1,2,3] Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]] Example 2: Input: nums = [0,1] Output: [[0,1],[1,0]] Example 3: Input: nums = [1] Output: [[1]]    Constraints:  1 <= nums.length <= 6 -10 <= nums[i] <= 10 All the integers of nums are unique.  

	# Explanation
	Here's an efficient solution to generate all possible permutations of an array of distinct integers:

*   **Backtracking:** Use a recursive backtracking approach to explore all possible arrangements of the numbers.
*   **Visited Set:** Maintain a set to keep track of which numbers have already been included in the current permutation. This avoids duplicates and ensures distinct permutations.
*   **Building Permutations:** Incrementally build permutations by adding numbers to the current permutation and recursively calling the function to add the remaining numbers.

*   **Runtime Complexity:** O(n!), where n is the number of elements in `nums`.
*   **Storage Complexity:** O(n!), mainly due to storing the resulting permutations.

	
	# Code
	```python
	def permute(nums):
    """
    Generates all possible permutations of a list of distinct integers.

    Args:
        nums: A list of distinct integers.

    Returns:
        A list of lists, where each inner list represents a permutation of the input list.
    """

    result = []
    n = len(nums)

    def backtrack(combination, used):
        if len(combination) == n:
            result.append(combination.copy())  # Append a copy to avoid modification
            return

        for num in nums:
            if num not in used:
                combination.append(num)
                used.add(num)
                backtrack(combination, used)
                used.remove(num)  # Backtrack: remove the last element
                combination.pop()  # Backtrack: remove the last element

    backtrack([], set())
    return result
	```