0741EasyArrayHash TableStringCounting

Signs Spellable From A Tile Tray

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Trains the technique from

LeetCode 1160Find Words That Can Be Formed by Characters

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 sign shop keeps one tray of loose letter tiles, given as the string tray: each character of tray is one physical tile, so a letter appearing three times in tray means the shop owns three tiles of it.

A sign whose text is label can be cut when the tray holds enough tiles to spell it: every letter of label needs its own tile, so a label using a letter twice needs two tiles of that letter. Deciding whether one label can be cut never uses up the tray — each label in labels is judged against the full tray, and the same label may appear in labels more than once, in which case it counts each time.

Return the total number of tiles the shop would place across all the labels it can cut, which is the sum of the lengths of those labels.

Examples

Example 1

Input: labels = ["glow", "gong", "log", "wool"], tray = "gowlno"
Output: 11

The tray stocks one `g`, two `o`, one `w`, one `l` and one `n`. `"glow"` uses four tiles and `"log"` uses three, and `"wool"` uses four including both `o` tiles, so those three labels contribute 4 + 3 + 4. `"gong"` needs two `g` tiles and the tray has one.

Example 2

Input: labels = ["pepper", "per"], tray = "prepe"
Output: 3

The tray stocks two `p`, one `r` and two `e`. `"pepper"` needs three `p` tiles, which the tray cannot supply. `"per"` needs one of each and contributes its three tiles.

Example 3

Input: labels = ["zz"], tray = "z"
Output: 0

The label wants two `z` tiles and the tray holds one, so no label can be cut and no tiles are placed.

Constraints

1 <= labels.length <= 1000
1 <= labels[i].length, tray.length <= 100
labels[i] and tray consist of 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 total_spellable_length(labels: list[str], tray: str) -> int:

Java

public int totalSpellableLength(String[] labels, String tray)

The intended approach

Time O(L + T) for total label length L and tray length T, plus a fixed 26-letter comparison per label.Space O(1): two 26-slot tallies.

Count how many tiles of each of the 26 letters the tray holds, once, before looking at any label. For a label, count its own letters the same way and cut it only when no letter is wanted more times than the tray stocks it. The tray tally is never decremented, which is what keeps every label judged against the whole tray. Add the length of each label that passes.

Pythonentry point: total_spellable_length

def total_spellable_length(labels, tray):
    # tally the tray once, then tally each label and compare letter by letter.
    # O(L + T) for total label length L and tray length T.
    stock = [0] * 26
    for ch in tray:
        stock[ord(ch) - 97] += 1
    total = 0
    for label in labels:
        need = [0] * 26
        for ch in label:
            need[ord(ch) - 97] += 1
        if all(need[k] <= stock[k] for k in range(26)):
            total += len(label)
    return total

The obvious version

For each label, copy the tray into a list of tiles and remove one matching tile per letter of the label, abandoning the label as soon as a needed tile is not in the pile. No tallies at all, just repeated searches through the pile.

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_total_spellable_length(labels, tray):
    """Physically hand out tiles: for each label take a fresh copy of the tray as
    a list and remove one matching tile per letter, giving up if one is missing."""
    total = 0
    for label in labels:
        pile = list(tray)
        ok = True
        for ch in label:
            if ch in pile:
                pile.remove(ch)
            else:
                ok = False
                break
        if ok:
            total += len(label)
    return total

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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