Smallest Palindromic Rearrangement I

• Count the frequency of each character in the input string s.
• Define a recursive backtracking function, generate(current_permutation, character_counts).
• Base Case: If the length of current_permutation equals the length of s:
  • Check if current_permutation is a palindrome.
  • If it is, compare it with the smallest palindrome found so far and update if the current one is smaller.
  • Return.
• Recursive Step: Iterate through all possible characters ('a' to 'z').
  • If a character's count is greater than zero:
    • Decrement its count.
    • Append the character to current_permutation.
    • Make a recursive call: generate(current_permutation, character_counts).
    • Backtrack: Remove the last character from current_permutation and restore the character's count.
• Initialize the process by calling the function with an empty permutation.
• The smallest palindrome found after exploring all possibilities is the answer.

The algorithm first needs a way to generate all unique permutations of the characters of s. This is typically done using a recursive backtracking function. The function would explore placing each available character at the current position in the permutation being built. To handle duplicate characters, we can use a frequency map of characters. Once a full permutation of length n is generated, we check if it's a palindrome. A string is a palindrome if it reads the same forwards and backwards. We maintain a variable, say smallestPalindrome, initialized to a lexicographically large value. Whenever we find a new palindrome, we compare it with smallestPalindrome and update it if the new one is smaller. After checking all possible permutations, smallestPalindrome will hold the answer. Given the constraints (s.length <= 10^5), this approach is computationally infeasible.

1// Note: This code is for demonstration and will cause a Time Limit Exceeded error.2class Solution {3    String smallestPalindrome = null;4 5    public String makeSmallestPalindrome(String s) {6        int[] counts = new int[26];7        for (char c : s.toCharArray()) {8            counts[c - 'a']++;9        }10        generatePermutations(new StringBuilder(), counts, s.length());11        return smallestPalindrome;12    }13 14    private void generatePermutations(StringBuilder current, int[] counts, int n) {15        if (current.length() == n) {16            String perm = current.toString();17            if (isPalindrome(perm)) {18                if (smallestPalindrome == null || perm.compareTo(smallestPalindrome) < 0) {19                    smallestPalindrome = perm;20                }21            }22            return;23        }24 25        for (int i = 0; i < 26; i++) {26            if (counts[i] > 0) {27                counts[i]--;28                current.append((char) ('a' + i));29                generatePermutations(current, counts, n);30                // Backtrack31                current.deleteCharAt(current.length() - 1);32                counts[i]++;33            }34        }35    }36 37    private boolean isPalindrome(String str) {38        int left = 0;39        int right = str.length() - 1;40        while (left < right) {41            if (str.charAt(left) != str.charAt(right)) {42                return false;43            }44            left++;45            right--;46        }47        return true;48    }49}