How do you solve Letter Tile Possibilities on LeetCode?

Letter Tile Possibilities (LeetCode #1079) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n!) time complexity and O(n) space complexity. Key insight: Using a frequency count helps avoid duplicates.

What is the brute force solution for Letter Tile Possibilities?

The brute force approach for Letter Tile Possibilities is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach generates all possible sequences of letters from the tiles. This involves exploring every combination of tiles, including duplicates, to count how many unique sequences can be formed. The optimal Optimal Solution approach improves this to O(n!).

What algorithm or data structure is used to solve Letter Tile Possibilities?

Letter Tile Possibilities uses the following concepts: Hash Table, String, Backtracking, Counting. The recommended patterns to study are: Backtracking, Hash Map.

What are the interview tips for Letter Tile Possibilities?

Explain your thought process clearly while coding. Discuss edge cases, like all tiles being the same. Practice writing clean and efficient code.

What are common mistakes when solving Letter Tile Possibilities?

Forgetting to check for duplicates when using characters. Not using a set to track unique sequences.

#1079

Letter Tile Possibilities

Medium

Hash TableStringBacktrackingCountingBacktrackingHash Map

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n!)
Space	O(1)	O(n)

💡

Intuition

Time O(n!)Space O(n)

The optimal solution uses backtracking with a frequency count of each character to avoid generating duplicate sequences. This significantly reduces the number of recursive calls and ensures we only consider unique characters.

⚙️

Algorithm

3 steps

1Step 1: Count the frequency of each character in the tiles.
2Step 2: Use a backtracking function that builds sequences based on the frequency count.
3Step 3: For each character, if it can be used (frequency > 0), decrease its frequency, add it to the current sequence, and recurse. Restore the frequency after recursion.

solution.py13 lines

1from collections import Counter
2
3def numTilePossibilities(tiles):
4    def backtrack(counter):
5        total = 0
6        for char in counter:
7            if counter[char] > 0:
8                counter[char] -= 1
9                total += 1 + backtrack(counter)
10                counter[char] += 1
11        return total
12
13    return backtrack(Counter(tiles))

ℹ

Complexity note: The time complexity is O(n!) because we explore all permutations of the characters, but we avoid duplicates by using a frequency count.

1Using a frequency count helps avoid duplicates.
2Backtracking allows us to explore all combinations efficiently.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.