Permutations Ii

To solve this coding challenge, we need to generate all unique permutations of a list of numbers that might contain duplicates. To achieve this, we can use a backtracking algorithm combined with a way to skip duplicates.

Explanation

The problem requires us to generate all unique permutations of an input list

nums

. The list may have duplicate elements, so we must ensure that the generated permutations are unique, even if

nums

contains duplicates. Here are the steps we'll take to solve this:

1. Backtracking Approach: We'll use backtracking to explore all permutations. Backtracking involves a recursive approach where we consider each element for the current position, swap elements, and move to the next position.
2. Skip Duplicates: To handle duplicates and avoid generating the same permutation multiple times, we use a set to track which elements have already been used at the current position. This prevents us from considering the same element more than once for the same position in the current recursive call.
3. Result Storage: We'll store each unique permutation in a result list. Once we reach a valid permutation (when the
   start
   index reaches the length of
   nums
   ), we add a copy of the current permutation to the result list.
4. Recursive Function: The main part of the algorithm is a recursive function
   backtrack(start)
   that iterates through the elements from the current position
   start
   to the end of the list. For each element, if it hasn't been used at the current position (checked using the set), we swap it to the current position, recursively call
   backtrack
   for the next position, and then swap back to restore the original list structure.

Step-by-Step Explanation and Detailed Steps in Pseudocode

5. Initialize the Result List
• We start by creating an empty list
  result
  to store all unique permutations.
6. Define the Backtracking Function
• Define a function
  backtrack(start)
  that will recursively build permutations.
• If
  start
  equals the length of
  nums
  , it means we have a complete permutation, so we append a copy of
  nums
  to
  result
  .
• Use a set
  seen
  to track elements that have been added to the current position to skip duplicates.
7. Iterate and Swap Elements
• Iterate through elements starting from the
  start
  index.
• For each element, if it is not in the
  seen
  set, add it to
  seen
  , swap it with the current
  start
  element, recursively call
  backtrack(start + 1)
  , and then swap back to restore the original list.
8. Return the Result
• Finally, after all recursive calls, return the
  result
  containing all unique permutations.

                                            
# Function to generate all unique permutations
function permuteUnique(nums):
    # List to hold all unique permutations
    result = []

    # Helper function for backtracking
    function backtrack(start):
        # If we have a complete permutation
        if start == length(nums):
            # Append a copy of the current permutation to result
            result.append(copy of nums)
            return

        # Set to track used elements at the current position
        seen = set()
        
        # Iterate through elements starting from the current position
        for i = start to length(nums) - 1:
            # If the element has been seen, skip it to prevent duplicate permutations
            if nums[i] in seen:
                continue
            
            # Add the element to the seen set
            seen.add(nums[i])
            
            # Swap the current element to the start position
            swap(nums, start, i)
            
            # Recursively build permutations with the next position
            backtrack(start + 1)
            
            # Swap back to restore the original list
            swap(nums, start, i)

    # Start backtracking from the first position
    backtrack(0)
    
    # Return the list of unique permutations
    return result

# Swap function to swap elements at two indices in nums
function swap(nums, i, j):
    temp = nums[i]
    nums[i] = nums[j]
    nums[j] = temp

                                        

This pseudocode outlines the overall structure of the solution, incorporating backtracking and the handling of duplicates to ensure all generated permutations are unique.