Introduction

In this blog post, we will explore how to solve the LeetCode problem "1079" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You have n tiles, where each tile has one letter tiles[i] printed on it. Return the number of possible non-empty sequences of letters you can make using the letters printed on those tiles. Example 1: Input: tiles = "AAB" Output: 8 Explanation: The possible sequences are "A", "B", "AA", "AB", "BA", "AAB", "ABA", "BAA". Example 2: Input: tiles = "AAABBC" Output: 188 Example 3: Input: tiles = "V" Output: 1 Constraints: 1 <= tiles.length <= 7 tiles consists of uppercase English letters.

Explanation

Here's a breakdown of the solution and the Python code:

• High-Level Approach:

  • Count the frequency of each letter in the input tiles.
  • Use recursion (Depth-First Search - DFS) to explore all possible sequences. The core idea is to try adding each available letter to the current sequence and recursively explore further sequences.
  • Handle duplicates effectively by using the counts of each letter and backtracking once a letter's count reaches zero.

• Complexity:

  • Runtime Complexity: O(N!), where N is the length of the input tiles. In the worst case, we explore all possible permutations.
  • Storage Complexity: O(N), mainly due to the recursion depth and the frequency counter.

Code

    from collections import Counter

def numTilePossibilities(tiles: str) -> int:
    """
    Calculates the number of possible non-empty sequences of letters
    that can be made using the letters printed on the given tiles.

    Args:
        tiles: A string representing the letters on the tiles.

    Returns:
        The number of possible non-empty sequences.
    """

    count = Counter(tiles)
    result = 0

    def dfs():
        nonlocal result
        for char in count:
            if count[char] > 0:
                result += 1
                count[char] -= 1
                dfs()
                count[char] += 1  # Backtrack

    dfs()
    return result