1168MediumHash TableStringBacktrackingCounting

Rows That Can Be Laid From the 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 bag holds tiles, each stamped with one uppercase letter, given by tiles. Tiles may be laid out in a row in any order, using anywhere from one of them up to all of them.

Return how many different non-empty strings can be laid out. Two rows count as the same string when they read alike.

Examples

Example 1

Input: tiles = "AB"
Output: 4

The rows are A, B, AB and BA.

Example 2

Input: tiles = "AA"
Output: 2

The two tiles read alike, so the only rows are A and AA.

Example 3

Input: tiles = "ABC"
Output: 15

Three rows of one tile, six of two tiles and six of all three.

Constraints

1 <= tiles.length <= 7
tiles holds only uppercase English letters

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 num_tile_possibilities(tiles: str) -> int:

Java

public int numTilePossibilities(String tiles)

The intended approach

Time O(n!) rows in the worst case, with a fixed cost each.Space O(a) for the tallies and O(n) for the recursion.

Two tiles stamped with the same letter are interchangeable, so what the bag really offers is a tally per letter. Building rows from tallies rather than from tiles is what stops the same string being counted once per way of picking which copies were used. So grow rows one letter at a time. At each step, try each letter that still has a copy left, count the row reached by adding it, and carry on growing from there before putting the copy back. Every non-empty row is reached exactly once, by its own letters in its own order, so the count comes out right with no set of strings to keep. The bag holds at most seven tiles, so the growing stops quickly.

Pythonentry point: num_tile_possibilities

def num_tile_possibilities(tiles):
    counts = {}
    for ch in tiles:
        counts[ch] = counts.get(ch, 0) + 1
    letters = sorted(counts)

    def walk():
        total = 0
        for ch in letters:
            if counts[ch] == 0:
                continue
            counts[ch] -= 1
            total += 1 + walk()
            counts[ch] += 1
        return total

    return walk()

The obvious version

Spell out every arrangement of every number of tiles, gather the strings they read into a set so repeats collapse, and report how many the set holds. Costs the tile count factorial in arrangements, most of which read alike.

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_num_tile_possibilities(tiles):
    from itertools import permutations
    seen = set()
    for size in range(1, len(tiles) + 1):
        for order in permutations(tiles, size):
            seen.add("".join(order))
    return len(seen)

How this problem was checked

Reference solution executed against 15 test cases — 15 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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