Introduction

In this blog post, we will explore how to solve the LeetCode problem "47" 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 a collection of numbers, nums, that might contain duplicates, return all possible unique permutations in any order. Example 1: Input: nums = [1,1,2] Output: [[1,1,2], [1,2,1], [2,1,1]] Example 2: Input: nums = [1,2,3] Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]] Constraints: 1 <= nums.length <= 8 -10 <= nums[i] <= 10

Explanation

• Backtracking with Pruning: The solution utilizes a backtracking algorithm to explore all possible permutations. To avoid generating duplicate permutations, it employs a pruning technique based on sorting the input and skipping identical numbers during the permutation construction process.
  • Frequency Counting: The solution creates a frequency count dictionary to efficiently manage the number of times each number appears in the initial array.
  • Recursive Helper: A recursive helper function constructs permutations by making choices at each step. The function only considers unique numbers to add to the current permutation.

• Time Complexity: O(n!): where n is the number of elements in the input array nums. In the worst case, the algorithm needs to explore all possible n! permutations. Space Complexity: O(n): where n is the number of elements in the input array nums. This space is used for the recursion stack and the storage of the generated permutations and the frequency count dictionary.

Code

    def permuteUnique(nums):
    """
    Generates all unique permutations of a list of numbers that may contain duplicates.

    Args:
        nums: A list of integers.

    Returns:
        A list of lists, where each inner list is a unique permutation of nums.
    """

    def backtrack(combination, remaining, permutations):
        if not remaining:
            permutations.append(combination.copy())
            return

        for num in remaining:
            new_remaining = remaining.copy()
            new_remaining.remove(num)
            backtrack(combination + [num], new_remaining, permutations)


    nums.sort()
    permutations = []

    def backtrack_optimized(combination, remaining_counts, permutations):
        if len(combination) == len(nums):
            permutations.append(combination.copy())
            return

        for num in remaining_counts:
            if remaining_counts[num] > 0:
                combination.append(num)
                remaining_counts[num] -= 1
                backtrack_optimized(combination, remaining_counts, permutations)
                remaining_counts[num] += 1
                combination.pop()


    remaining_counts = {}
    for num in nums:
        remaining_counts[num] = remaining_counts.get(num, 0) + 1

    permutations = []
    backtrack_optimized([], remaining_counts, permutations)
    return permutations