Introduction

In this blog post, we will explore how to solve the LeetCode problem "1138" 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

On an alphabet board, we start at position (0, 0), corresponding to character board[0][0]. Here, board = ["abcde", "fghij", "klmno", "pqrst", "uvwxy", "z"], as shown in the diagram below. We may make the following moves: 'U' moves our position up one row, if the position exists on the board; 'D' moves our position down one row, if the position exists on the board; 'L' moves our position left one column, if the position exists on the board; 'R' moves our position right one column, if the position exists on the board; '!' adds the character board[r][c] at our current position (r, c) to the answer. (Here, the only positions that exist on the board are positions with letters on them.) Return a sequence of moves that makes our answer equal to target in the minimum number of moves. You may return any path that does so. Example 1: Input: target = "leet" Output: "DDR!UURRR!!DDD!" Example 2: Input: target = "code" Output: "RR!DDRR!UUL!R!" Constraints: 1 <= target.length <= 100 target consists only of English lowercase letters.

Explanation

Here's a breakdown of the approach, complexity, and the Python code:

• High-Level Approach:

  • Create a dictionary (or similar data structure) to store the coordinates (row, col) of each letter in the alphabet board for quick lookup.
  • Iterate through the target string, for each character compute the moves from current position to next character using row and col differences.
  • Build the move string by appending directional characters ('U', 'D', 'L', 'R') and the '!' character for selecting.

• Complexity:

  • Runtime Complexity: O(N), where N is the length of the target string. The coordinate lookup and move generation for each character is done in constant time.
  • Storage Complexity: O(1), the size of the coordinate dictionary is constant since the alphabet size is fixed.

Code

    def alphabetBoardPath(target: str) -> str:
    """
    Finds the shortest path on an alphabet board to spell out the target string.
    """

    board = ["abcde", "fghij", "klmno", "pqrst", "uvwxy", "z"]
    pos = {}
    for r in range(len(board)):
        for c in range(len(board[r])):
            pos[board[r][c]] = (r, c)

    curr_row, curr_col = 0, 0
    path = ""

    for char in target:
        target_row, target_col = pos[char]
        row_diff = target_row - curr_row
        col_diff = target_col - curr_col

        # Move up/down first if moving to 'z'
        if char == 'z':
            if row_diff > 0:
                path += 'D' * row_diff
            if col_diff > 0:
                path += 'R' * col_diff
            if row_diff < 0:
                path += 'U' * abs(row_diff)
            if col_diff < 0:
                path += 'L' * abs(col_diff)
        else:
            if row_diff < 0:
                path += 'U' * abs(row_diff)
            if col_diff < 0:
                path += 'L' * abs(col_diff)

            if row_diff > 0:
                path += 'D' * row_diff
            if col_diff > 0:
                path += 'R' * col_diff

        path += '!'
        curr_row, curr_col = target_row, target_col

    return path