0335MediumStringSortingCounting Sort

Smallest Mirror Word From The Same Beads

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Trains the technique from

LeetCode 3517Smallest Palindromic Rearrangement I

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 bracelet is threaded with lettered beads, written out as the string s. The bracelet already reads the same forwards and backwards.

A jeweller is going to unthread every bead and thread them all back on in some order. The rethreaded bracelet must use exactly the beads that came off, all of them and nothing else, and it must again read the same forwards and backwards.

Of all the rethreadings that qualify, return the one that comes first alphabetically.

Examples

Example 1

Input: s = "dcabbacd"
Output: "abcddcba"

The beads are two each of a, b, c and d. The word `abcddcba` uses all eight of them and reads the same in both directions.

Example 2

Input: s = "cabac"
Output: "acbca"

There are two a, two c and a single b. The returned word reads the same in both directions and uses every bead.

Example 3

Input: s = "aabbaa"
Output: "aabbaa"

Four a beads and two b beads are already threaded in the alphabetically first qualifying order, so the bracelet comes back unchanged.

Example 4

Input: s = "cbabc"
Output: "bcacb"

Two b beads, two c beads and one a bead. The word `bcacb` uses all five and reads the same in both directions.

Constraints

1 <= s.length <= 10^5
s consists of lowercase English letters.
s reads the same forwards and backwards.

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_mirror_word(beads: str) -> str:

Java

public String smallestMirrorWord(String beads)

The intended approach

Time O(n + 26)Space O(n) for the returned word plus O(26) for the tally

Tally the 26 letter counts. At most one letter can have an odd count and that letter has to sit at the centre; every other bead pairs up, one bead into each arm. Emitting half of every count in alphabetical order gives the opening arm, and the closing arm is that arm read backwards.

Pythonentry point: smallest_mirror_word

def smallest_mirror_word(beads):
    # Every letter occurs an even number of times except at most one, so half of
    # each count fills the opening arm; sorting that arm ascending makes the whole
    # word as small as it can be, and the closing arm is forced to be its mirror.
    # Time O(n + 26). Space O(n) for the word that is returned.
    tally = [0] * 26
    base = ord("a")
    for ch in beads:
        tally[ord(ch) - base] += 1
    arm = []
    middle = ""
    for i in range(26):
        if tally[i] % 2 == 1:
            middle = chr(base + i)
        if tally[i] // 2:
            arm.append(chr(base + i) * (tally[i] // 2))
    head = "".join(arm)
    return head + middle + head[::-1]

The obvious version

Enumerate every distinct rethreading, keep the ones that read the same in both directions, and take the alphabetically first.

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_mirror_word(beads):
    from itertools import permutations
    best = None
    for p in set(permutations(beads)):
        t = "".join(p)
        if t == t[::-1]:
            if best is None or t < best:
                best = t
    return best

How this problem was checked

Reference solution executed against 9 test cases — 9 passed
4 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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