0147EasyHash TableStringCounting

Spell It With 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 signmaker's kit holds a jumble of engraved letter tiles. The string tiles lists what the kit contains, one lowercase character per tile, in no particular arrangement. A customer orders the label word, which has to be laid along a display rail exactly as written.

Every tile is a physical piece, so a tile placed on the rail is no longer in the kit; a label needing the same letter three times needs three separate tiles carrying it. Tiles left over in the kit are fine and the order they sit in never matters.

Report true when the kit can cover the whole label, and false when it runs short.

Examples

Example 1

Input: word = "kite", tiles = "tickle"
Output: true

The kit carries a k, an i, a t and an e, which is everything the label asks for.

Example 2

Input: word = "peer", tiles = "prime"
Output: false

The label wants two e tiles and the kit only holds one, so it runs short.

Example 3

Input: word = "cat", tiles = "tactic"
Output: true

Three of the six tiles cover the label and the leftovers stay in the kit.

Constraints

1 <= word.length, tiles.length <= 10^5
word and tiles hold lowercase 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 can_construct(word: str, tiles: str) -> bool:

Java

public boolean canConstruct(String word, String tiles)

The intended approach

Time O(n + m)Space O(1)

Tally how many tiles of each letter the kit holds. Walk the label once, spending a tile per character, and fail the moment a letter's tally is already exhausted. Only the twenty-six lowercase letters are ever tallied, so the extra room is fixed.

Pythonentry point: can_construct

def can_construct(word, tiles):
    stock = {}
    for ch in tiles:
        stock[ch] = stock.get(ch, 0) + 1
    for ch in word:
        left = stock.get(ch, 0)
        if left == 0:
            return False
        stock[ch] = left - 1
    return True

The obvious version

Keep the kit as a list of tiles and, for each character of the label, scan the list for a matching tile and physically remove it. Quadratic and it counts nothing.

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_can_lay_out(word, tiles):
    kit = list(tiles)
    for ch in word:
        found = -1
        for i in range(len(kit)):
            if kit[i] == ch:
                found = i
                break
        if found < 0:
            return False
        kit.pop(found)
    return True

How this problem was checked

Reference solution executed against 12 test cases — 12 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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