0919HardHash TableMathStringCombinatoricsCounting

The k-th Mirrored Rearrangement

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Trains the technique from

LeetCode 3518Smallest Palindromic Rearrangement II

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A tag reads s, and the tag is already the same read either way.

Consider every rearrangement of the tag's letters that also reads the same either way, counting each distinct rearrangement once. List them in dictionary order.

Return the k-th one, or the empty string when fewer than k exist.

Examples

Example 1

Input: s = "abcba", k = 4
Output: ""

The front half holds one a and one b, so the halves in order are "ab" and "ba", giving only two rearrangements. There is no fourth, so nothing is returned.

Example 2

Input: s = "abcba", k = 1
Output: "abcba"

The smallest front half is "ab", the middle letter is the odd one out, and mirroring the half gives back the tag itself.

Example 3

Input: s = "aabbaa", k = 3
Output: "baaaab"

The front half holds two a and one b, so the halves in order are "aab", "aba" and "baa"; the third mirrors into this tag.

Constraints

1 <= s.length <= 10^4
s consists of lowercase English letters only
s reads the same either way
1 <= k <= 10^6

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def smallest_palindrome(s: str, k: int) -> str:

Java

public String smallestPalindrome(String s, int k)

The intended approach

Time O(n * A) for a tag of length n over an alphabet of size A, each position costing one pass over the alphabet with a capped count.Space O(A) for the letter counts.

A rearrangement that reads the same either way is settled entirely by its front half, since the back half is the front reversed and the middle letter, if the tag has odd length, is forced to be the one letter with an odd count. Dictionary order on the whole tag follows dictionary order on the front halves, so the job reduces to finding the `k`-th arrangement of the half's letters. Build that half one position at a time. For each letter in alphabetical order, work out how many arrangements begin with it, which is the number of distinct orderings of the letters that would remain. If `k` is larger than that, subtract it and move to the next letter; otherwise fix this letter and carry on to the next position. Counting the orderings exactly would overflow on a long tag, so cap every count at `k`, which is all that is ever compared against, and the arithmetic stays small.

Pythonentry point: smallest_palindrome

def smallest_palindrome(s, k):
    CAP = 2 * 10 ** 6
    counts = [0] * 26
    for ch in s:
        counts[ord(ch) - 97] += 1
    middle = ""
    for c in range(26):
        if counts[c] % 2 == 1:
            middle = chr(97 + c)
        counts[c] //= 2
    half_len = 0
    for c in range(26):
        half_len += counts[c]

    def orderings(left):
        # distinct arrangements of the remaining letters, capped so it stays small
        total = 0
        for c in range(26):
            total += left[c]
        out = 1
        used = 0
        for c in range(26):
            take = left[c]
            if take == 0:
                continue
            # multiply by (total - used) choose take, capping as we go
            for step in range(1, take + 1):
                out = out * (total - used - take + step) // step
                if out > CAP:
                    return CAP
            used += take
        return out

    if orderings(counts) < k:
        return ""
    half = []
    for _ in range(half_len):
        for c in range(26):
            if counts[c] == 0:
                continue
            counts[c] -= 1
            here = orderings(counts)
            if here >= k:
                half.append(chr(97 + c))
                break
            k -= here
            counts[c] += 1
    front = "".join(half)
    return front + middle + front[::-1]

The obvious version

Build every arrangement of the front half outright, keep the distinct ones, sort them, and read off the k-th, mirroring it back into a full tag. Factorial in the half's length and it counts nothing in advance.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_smallest_palindrome(s, k):
    import itertools
    counts = {}
    for ch in s:
        counts[ch] = counts.get(ch, 0) + 1
    middle = ""
    half_letters = []
    for ch in sorted(counts):
        if counts[ch] % 2 == 1:
            middle = ch
        half_letters.extend([ch] * (counts[ch] // 2))
    seen = set()
    for order in itertools.permutations(half_letters):
        seen.add("".join(order))
    ranked = sorted(seen)
    if k > len(ranked):
        return ""
    front = ranked[k - 1]
    return front + middle + front[::-1]

How this problem was checked

Reference solution executed against 13 test cases — 13 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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