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, 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

    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