Solution Explanation

This problem asks us to find all unique palindromic permutations of a given string s. A palindromic permutation is a permutation of the characters in the string that forms a palindrome (reads the same forwards and backward).

The solution uses a backtracking approach with a helper function dfs. Let's break down the process:

1. Character Count: First, we count the occurrences of each character in the input string s using a hash map (dictionary in Python, unordered_map in C++, map in Go, and an array in Java).

2. Mid Character: We check if the string can even form a palindrome. A string has a palindromic permutation only if at most one character has an odd count. We identify this character (if it exists) as the mid character which will be placed in the center of the palindrome. If more than one character has an odd count, it's impossible to form a palindrome, and we return an empty list.

3. Backtracking (dfs): The core logic resides in the recursive dfs function:

   • Base Case: If the length of the current string t being built equals the length of the input string s, it means we've formed a palindrome. We add it to the ans list (which stores the results).
   • Recursive Step: We iterate through the characters and their counts. If a character has a count greater than 1, we can use it to extend the palindrome by adding it to both the beginning and the end of the current string t (c + t + c). We then recursively call dfs with the updated string and counts, and after the recursive call, we backtrack by restoring the character count, ensuring we explore all possibilities.

4. Return: Finally, we return the ans list containing all unique palindromic permutations.

Time and Space Complexity Analysis

• Time Complexity: The time complexity is dominated by the backtracking process. In the worst case, we might explore all possible permutations. The number of permutations can be significantly large (though bounded by the length of the input string, it can still be substantial for longer strings). Therefore, the time complexity is roughly O(N!), where N is the length of the string (s). However, it's not exactly N! because the algorithm prunes branches when there aren't enough characters left to form a palindrome. A tighter bound might be hard to derive analytically but it would still be exponential in nature.

• Space Complexity: The space complexity is determined by the following:

  • The hash map (or array) to store character counts: O(1) as the alphabet size (26 lowercase letters) is constant.
  • The recursion stack during the backtracking process: In the worst case, the depth of the recursion is O(N/2), where N is the length of the string. This is because we only add characters in pairs (except for the potential single mid character).
  • The ans list to store the generated palindromes: This list can grow to contain many palindromes in the worst case. In the worst case scenario, the number of palindromes is exponential as well. Therefore, the overall space complexity can be considered as O(N) + number of palindromes, which can be considered O(N) + exponential in the worst case.