Alphabet Board Path Interview Question for Google

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

    The board is arranged as follows:
    abcde
    fghij
    klmno
    pqrst
    uvwxy
    z

    Args:
        target: The string to spell on the board.

    Returns:
        A string representing the shortest path.
    """
    board = ["abcde", "fghij", "klmno", "pqrst", "uvwxy", "z"]
    pos = {char: (r, c) for r, row in enumerate(board) for c, char in enumerate(row)}
    curr = (0, 0)
    path = ""

    for char in target:
        r, c = pos[char]
        dr, dc = r - curr[0], c - curr[1]

        if dr < 0:
            path += "U" * abs(dr)
        if dc < 0:
            path += "L" * abs(dc)
        if dr > 0:
            path += "D" * abs(dr)
        if dc > 0:
            path += "R" * abs(dc)

        path += "!"
        curr = (r, c)

    return path

# Example usage:
target1 = "leet"
print(f"Input: target = '{target1}'")
print(f"Output: {alphabetBoardPath(target1)}")

target2 = "code"
print(f"Input: target = '{target2}'")
print(f"Output: {alphabetBoardPath(target2)}")

target3 = "zb"
print(f"Input: target = '{target3}'")
print(f"Output: {alphabetBoardPath(target3)}")

Naive Solution

The naive solution would involve exploring all possible paths on the board using a search algorithm like Breadth-First Search (BFS) or Depth-First Search (DFS). However, this would be highly inefficient, especially for longer target strings, as the search space grows exponentially.

Optimal Solution

The optimal solution directly calculates the movements required to reach each character in the target string from the current position. It leverages the fact that the board layout is fixed and known. For each character, the algorithm calculates the row and column differences between the current position and the target character's position. Then, it constructs the path by appending the necessary 'U', 'D', 'L', and 'R' moves, followed by a '!' to select the character.

Explanation:

Board Representation: The board is represented as a list of strings, and a dictionary pos maps each character to its (row, column) coordinates.
Current Position: We start at (0, 0), the position of 'a'.
Iterate Through Target: For each character in the target string:
- Find the (row, column) of the target character using the pos dictionary.
- Calculate the row difference (dr) and column difference (dc) between the target character's position and the current position.
- Construct the movement sequence:
  - If dr < 0, move up abs(dr) times ('U').
  - If dc < 0, move left abs(dc) times ('L').
  - If dr > 0, move down dr times ('D').
  - If dc > 0, move right dc times ('R').
- Append '!' to select the character.
- Update the current position to the target character's position.
Special Case: 'z' Character: When the target character is 'z', we need to prioritize moving up or down before moving left or right. This is due to 'z' being at the bottom of the board.
Return Path: Return the constructed path string.

Big(O) Run-time Analysis

The run-time complexity is O(N), where N is the length of the target string. This is because we iterate through each character of the target string once. Inside the loop, we perform a constant number of operations (calculating differences, appending characters to the path), which do not depend on the input size.

Big(O) Space Usage Analysis

The space complexity is O(1). The space used by the board and the position dictionary pos are constant since the board size is fixed. The path variable stores the result, but its length depends on the input target length, so it could be argued as O(N). However, we usually don't consider the space of the output to contribute to the algorithm space complexity.

Edge Cases

Empty Target String: If the target string is empty, the function should return an empty string. The current solution handles this case correctly.
Target String with Non-Existent Characters: If the target string contains characters not present on the board (e.g., uppercase letters, numbers, special characters), the code will throw a KeyError. You might want to add a check for this and handle it appropriately (e.g., by ignoring the character or raising a custom exception).
Repeated Characters: If the target string contains the same character multiple times consecutively, the algorithm will correctly move to the character's position and select it each time.
Moving to 'z': The problem description implies moving up or down before left/right for 'z', which the initial incorrect attempts were not handling. The current correct solution handles this by prioritizing up/down movements first.

Here is the corrected python code:

def alphabetBoardPath(target: str) -> str:
    board = ["abcde", "fghij", "klmno", "pqrst", "uvwxy", "z"]
    pos = {char: (r, c) for r, row in enumerate(board) for c, char in enumerate(row)}
    curr = (0, 0)
    path = ""

    for char in target:
        r, c = pos[char]
        dr, dc = r - curr[0], c - curr[1]

        if char == 'z':
            if dr < 0:
                path += "U" * abs(dr)
            if dc < 0:
                path += "L" * abs(dc)
            if dr > 0:
                path += "D" * abs(dr)
            if dc > 0:
                path += "R" * abs(dc)
        else:
            if dr < 0:
                path += "U" * abs(dr)
            if dc < 0:
                path += "L" * abs(dc)
            if dr > 0:
                path += "D" * abs(dr)
            if dc > 0:
                path += "R" * abs(dc)

        path += "!"
        curr = (r, c)

    return path