Solution Explanation for Letter Tile Possibilities

This problem asks to find the number of unique non-empty sequences that can be formed using the given tiles. The solution uses a backtracking approach to explore all possible sequences.

Approach:

The core idea is to use a depth-first search (DFS) or backtracking algorithm. We maintain a count of each letter's frequency. In each recursive call, we try to include each letter present (whose count is greater than 0). After including a letter, we recursively explore further possibilities and then backtrack (restore the count) to explore other combinations.

Algorithm:

1. Count Letter Frequencies: First, we count the frequency of each letter in the input string tiles using a hash map (or array in the provided code examples).

2. Depth-First Search (DFS):

   • The dfs function takes the frequency counter as input.
   • It iterates through each letter. If a letter has a count greater than 0:
     • It increments the total count of possible sequences (ans).
     • It decrements the count of that letter.
     • It recursively calls dfs to explore sequences starting with the selected letter.
     • It increments back the count of the letter (backtracking step). This is crucial for exploring all possible combinations.

3. Return Total Count: The dfs function returns the total number of possible sequences found.

Time Complexity Analysis:

The time complexity is difficult to express precisely, but it's bounded by the number of possible permutations of the tiles. In the worst case (all letters are distinct), the complexity would be O(n!), where n is the length of tiles. However, because there can be duplicate letters, the actual number of permutations is significantly less than n!. The complexity is closer to the number of unique permutations considering the frequency of each letter.

Space Complexity Analysis:

The space complexity is dominated by the recursion depth of the DFS. In the worst case (all letters are distinct), the depth can be O(n). Additionally, the space used for the frequency counter is O(1) (since the number of unique characters is limited to 26). Therefore, the overall space complexity is O(n) due to the recursion stack.

Code Explanation (Python):

from collections import Counter
 
class Solution:
    def numTilePossibilities(self, tiles: str) -> int:
        def dfs(cnt: Counter) -> int:
            ans = 0
            for i, x in cnt.items():
                if x > 0:
                    ans += 1  # Count the current sequence
                    cnt[i] -= 1 # Include letter
                    ans += dfs(cnt) # Explore further
                    cnt[i] += 1 # Backtrack: remove the letter
            return ans
 
        cnt = Counter(tiles) # Count initial letter frequencies
        return dfs(cnt)      # Start the backtracking search
 

The other language solutions follow the same logic, only differing in syntax and data structure implementations. The core algorithm remains identical across all languages.