Letter Tile Possibilities

Letter Tile Possibilities

What is this problem about?

The "Letter Tile Possibilities interview question" involves a set of tiles, each with a letter printed on it. You need to find the total number of non-empty sequences of letters you can form using these tiles. Unlike a standard permutation problem, you don't have to use all tiles, and tiles can be identical. This "Letter Tile Possibilities coding problem" is a counting challenge that combines frequency analysis with recursion.

Why is this asked in interviews?

This problem is asked to test a candidate's ability to handle permutations of multisets using the "Backtracking interview pattern". Companies like Google and Oracle use it to see if you can avoid duplicate sequences by using a "Counting interview pattern" (frequency map) instead of just permuting the raw indices.

Algorithmic pattern used

The best way to solve this is to count the frequency of each letter first. Then, use a recursive function that tries to pick one available letter from the frequency map to start a sequence. For each letter picked, increment the total count and then recurse with the updated frequency map. This automatically handles duplicate letters and generates sequences of all possible lengths.

Example explanation

Tiles: "AAB"

1. Frequency: {A: 2, B: 1}
2. Pick 'A': Seq "A". Freq {A: 1, B: 1}. Recurse...
   • Pick 'A': Seq "AA". Freq {A: 0, B: 1}. Recurse -> "AAB".
   • Pick 'B': Seq "AB". Freq {A: 1, B: 0}. Recurse -> "ABA".
3. Pick 'B': Seq "B". Freq {A: 2, B: 0}. Recurse...
   • Pick 'A': Seq "BA". Freq {A: 1, B: 0}. Recurse -> "BAA". Total unique sequences: "A", "AA", "AAB", "AB", "ABA", "B", "BA", "BAA" (Total 8).

Common mistakes candidates make

• Generating all permutations and using a Set: This works but is very slow and memory-intensive for larger inputs.
• Not handling duplicates: Counting "A1A2" and "A2A1" as different sequences when they both look like "AA".
• Missing short sequences: Only counting sequences that use all the tiles.

Interview preparation tip

When a problem involves permutations of items with duplicates, always think about using a frequency map (or a sorted array) to manage your choices. It's much more efficient than using a Set to filter out duplicates at the end.