0443EasyHash TableStringGreedy

Longest Lintel Row of Carved Tiles

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Trains the technique from

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 mason keeps carved letter tiles loose in a crate. crate gives the carving on each tile, in the order the tiles come out of the crate. Carvings are cut with either an upper-case punch or a lower-case punch, and the two are different tiles: a tile carved S will not pass for a tile carved s.

The mason lays some of the tiles in one row along a lintel. The tiles may be laid in any order, any number of them may be left in the crate, and no tile can be used twice. The row must read the same scanned left to right as scanned right to left.

Return the largest number of tiles such a row can contain.

Examples

Example 1

Input: crate = "ledgerledger"
Output: 12

All twelve tiles go into the row, for example laid as `ledgerregdel`, which reads the same from either end.

Example 2

Input: crate = "Zz"
Output: 1

The two tiles carry different carvings, so the row holds a single tile, and one tile reads the same either way.

Example 3

Input: crate = "ppqr"
Output: 3

A row of three such as `pqp` uses both `p` tiles and one of the remaining two; the last tile stays in the crate.

Example 4

Input: crate = "granitegranit"
Output: 13

Thirteen of the tiles form a row, for instance `granitetinarg`.

Constraints

1 <= crate.length <= 2000
crate consists of lower-case and/or upper-case English letters only.

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 longest_band(crate: str) -> int:

Java

public int longestBand(String crate)

The intended approach

Time O(n) for one pass over the crate plus a scan of at most fifty-two tallies.Space O(1): the tally holds at most fifty-two entries however large the crate is.

Tally the tiles by carving, counting the two cases of a letter as separate carvings. Two tiles of one carving can face each other across the centre of the row, so each carving hands over twice its halved tally, and if any carving has a tile spare after that, exactly one spare tile can be set at the centre. Tally, then sum, then add the single centre tile if one is available.

Pythonentry point: longest_band

def longest_band(crate):
    # Tally the tiles per carving. Two tiles of the same carving can face each
    # other across the middle of the row, so each carving contributes twice its
    # halved tally. If any carving has a tile left over after that pairing, one
    # single tile can sit dead centre, and only one such tile can ever be placed.
    # Upper and lower case are tallied separately because they are different
    # carvings.
    # O(n) time, O(1) space: at most fifty-two tallies whatever the crate holds.
    tally = {}
    for ch in crate:
        tally[ch] = tally.get(ch, 0) + 1
    laid = 0
    spare = False
    for count in tally.values():
        laid += (count // 2) * 2
        if count % 2:
            spare = True
    if spare:
        laid += 1
    return laid

The obvious version

Act the pairing out on a working list: repeatedly look for two tiles carrying the same carving, pull both out and credit two tiles to the row. When no matching pair is left, credit one more tile if anything remains.

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_longest_band(crate):
    remaining = list(crate)
    laid = 0
    at = 0
    while at < len(remaining):
        carving = remaining[at]
        partner = -1
        for other in range(at + 1, len(remaining)):
            if remaining[other] == carving:
                partner = other
                break
        if partner >= 0:
            del remaining[partner]
            del remaining[at]
            laid += 2
        else:
            at += 1
    if remaining:
        laid += 1
    return laid

How this problem was checked

Reference solution executed against 12 test cases — 12 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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