# Solving Leetcode Interviews in Seconds with AI: Alphabet Board Path


	# 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](https://www.chatmagic.app), 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
	```python
	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
	```