Problem Description

You are given an alphabet board arranged as follows:

• Row 0: "abcde"
• Row 1: "fghij"
• Row 2: "klmno"
• Row 3: "pqrst"
• Row 4: "uvwxy"
• Row 5: "z"

Starting at position (0, 0) which corresponds to the letter 'a', you need to spell out a given target string by moving on the board and selecting letters.

You can perform the following moves:

• 'U' - move up one row (if valid)
• 'D' - move down one row (if valid)
• 'L' - move left one column (if valid)
• 'R' - move right one column (if valid)
• '!' - select the current letter at your position and add it to your answer

Given a target string, return a sequence of moves that spells out the target. The sequence should use the minimum number of moves, but any valid path that achieves this is acceptable.

For example, to spell "leet":

• Start at 'a' (0,0)
• Move to 'l' at position (2,2): need to go down 2 and right 2, then select with '!'
• Move to 'e' at position (0,4): need to go up 2 and right 2, then select with '!'
• Stay at 'e' and select again with '!'
• Move to 't' at position (3,4): need to go down 3, then select with '!'

The solution handles this by calculating the target position for each character using the formula: row = (ascii_value - 'a') // 5 and column = (ascii_value - 'a') % 5. It then moves from the current position to the target position following a specific order (left, up, right, down) to avoid invalid moves, especially when dealing with 'z' which is alone on the last row.

Intuition

The key insight is that we need to find the shortest path from our current position to each target character. Since the board is arranged in a grid pattern with letters in alphabetical order, we can easily calculate any letter's position using math: the row is (letter_value // 5) and column is (letter_value % 5), where letter_value = ord(character) - ord('a').

For moving between positions, the shortest path is simply the Manhattan distance - we need to move horizontally by |current_col - target_col| steps and vertically by |current_row - target_row| steps. The total moves will always be the sum of these two distances regardless of the order we take them.

However, there's a crucial edge case: the letter 'z' sits alone on row 5. This creates a problem because positions (5,1), (5,2), (5,3), and (5,4) don't exist on the board. If we're not careful about our movement order, we might try to move through these invalid positions.

The solution cleverly addresses this by enforcing a specific movement order: Left → Up → Right → Down. This order ensures we never move through invalid positions. Here's why:

• When moving to 'z' at position (5,0): By moving left first and up second, we ensure we reach column 0 before descending to row 5. Then moving right (nothing to do since 'z' is at column 0) and down gets us safely to 'z'.

• When moving from 'z' to any other letter: By moving left first (nothing to do since we're already at column 0) and up second, we immediately leave row 5 before attempting any rightward movement, avoiding the invalid positions.

This movement order works for all other transitions as well since the rest of the board is fully populated, making any path valid. The specific order just guarantees we handle the 'z' edge case correctly while still achieving the minimum number of moves.

Solution Approach

The implementation follows a straightforward simulation approach where we track our current position and move to each target character one by one.

Algorithm Steps:

1. Initialize position: Start at (i=0, j=0) representing position 'a' on the board.

2. Process each character: For each character c in the target string:

   • Calculate the target position using mathematical mapping:
     • v = ord(c) - ord("a") gives us the character's position in the alphabet (0-25)
     • x = v // 5 gives us the target row
     • y = v % 5 gives us the target column

3. Move to target position: Execute movements in the specific order (Left → Up → Right → Down):

   • Move Left: While j > y, decrement j and append "L" to the result
   • Move Up: While i > x, decrement i and append "U" to the result
   • Move Right: While j < y, increment j and append "R" to the result
   • Move Down: While i < x, increment i and append "D" to the result

4. Select character: After reaching the target position, append "!" to select the current character.

5. Build result: Join all the movement commands into a single string.

Why this order works:

The movement order is critical for handling the edge case of 'z'. Consider two scenarios:

• Moving to 'z' from any letter: We first move left to column 0, then up if needed, ensuring we're at column 0 before moving down to row 5. This avoids invalid positions like (5,2).

• Moving from 'z' to any letter: We're at (5,0). Moving left does nothing, then we move up to leave row 5, then right to the target column. This ensures we exit row 5 before attempting any rightward movement.

Time Complexity: O(n * m) where n is the length of the target string and m is the maximum Manhattan distance between any two characters (at most 9 moves between 'a' and 'z').

Space Complexity: O(n * m) for storing the result string with all movement commands.

Example Walkthrough

Let's trace through spelling the word "za" to demonstrate the solution approach, particularly highlighting the edge case handling.

Initial Setup:

• Start at position (0, 0) which is letter 'a'
• Target string: "za"

Step 1: Move to 'z'

• Current position: (0, 0)
• Calculate target position for 'z':
  • v = ord('z') - ord('a') = 25
  • x = 25 // 5 = 5 (target row)
  • y = 25 % 5 = 0 (target column)
• Target position: (5, 0)

Now apply movements in order (Left → Up → Right → Down):

• Left: Current j=0, target y=0. No movement needed.
• Up: Current i=0, target x=5. No movement needed (we're going down, not up).
• Right: Current j=0, target y=0. No movement needed.
• Down: Current i=0, target x=5. Move down 5 times: "DDDDD"
• Select: Add "!"

Result so far: "DDDDD!"

Step 2: Move to 'a'

• Current position: (5, 0) - we're at 'z'
• Calculate target position for 'a':
  • v = ord('a') - ord('a') = 0
  • x = 0 // 5 = 0 (target row)
  • y = 0 % 5 = 0 (target column)
• Target position: (0, 0)

Apply movements in order (Left → Up → Right → Down):

• Left: Current j=0, target y=0. No movement needed.
• Up: Current i=5, target x=0. Move up 5 times: "UUUUU"
• Right: Current j=0, target y=0. No movement needed.
• Down: Current i=0, target x=0. No movement needed.
• Select: Add "!"

Final result: "DDDDD!UUUUU!"

Why the order matters: Notice when moving from 'z' at (5,0) to 'a', we move UP first before any potential RIGHT movements. If we had tried to move right first from position (5,0), we would have attempted to reach invalid positions like (5,1) or (5,2) which don't exist on the board. By moving up first, we safely leave row 5 before any horizontal movements, ensuring we only traverse valid positions.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n), where n is the length of the target string. The algorithm iterates through each character in the target string exactly once. For each character, it performs a constant amount of work - calculating the target position (x, y) takes O(1) time, and moving from the current position to the target position takes at most O(1) time since the board has fixed dimensions (6×5 grid). The maximum distance between any two positions on the board is bounded by a constant (at most 5 moves vertically and 4 moves horizontally), so the inner while loops execute a constant number of times for each character.

Space Complexity: O(1) if we exclude the space used for the output string. The algorithm uses only a fixed number of variables (i, j, v, x, y, c) regardless of the input size. The ans list grows proportionally to store the result, but as mentioned in the reference, we ignore the space consumption of the answer itself when analyzing space complexity.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall: Invalid Board Access When Moving to/from 'z'

The most critical pitfall in this problem is not handling the special case of 'z' correctly. The letter 'z' is located at position (5, 0), making it the only character in row 5. If you attempt to move horizontally while on row 5 to any column other than 0, or if you try to reach row 5 at any column other than 0, you'll access an invalid position on the board.

Problematic Approach:

# WRONG: Moving in arbitrary order without considering 'z'
def alphabetBoardPath(self, target: str) -> str:
    i, j = 0, 0
    result = []
  
    for c in target:
        x = (ord(c) - ord('a')) // 5
        y = (ord(c) - ord('a')) % 5
      
        # This can fail when moving to/from 'z'
        while i < x: 
            i += 1
            result.append('D')
        while i > x:
            i -= 1
            result.append('U')
        while j < y:
            j += 1
            result.append('R')
        while j > y:
            j -= 1
            result.append('L')
      
        result.append('!')
    return ''.join(result)

Example of Failure:

• Moving from 'y' (4, 4) to 'z' (5, 0):
  • If we move down first: (4,4) → (5,4) - INVALID! Row 5 only has column 0
  • We must move left to column 0 first, then down

Solution: The correct approach uses a specific movement order: Left → Up → Right → Down

This order ensures:

1. When moving TO 'z': We move left to column 0 first, then down to row 5
2. When moving FROM 'z': We move up first to leave row 5, then right to the target column

Correct Implementation Pattern:

# Move in this specific order to handle 'z' safely
while current_col > target_col:  # Left first
    current_col -= 1
    result_path.append('L')

while current_row > target_row:  # Up second
    current_row -= 1
    result_path.append('U')

while current_col < target_col:  # Right third
    current_col += 1
    result_path.append('R')

while current_row < target_row:  # Down last
    current_row += 1
    result_path.append('D')

This movement order guarantees we never attempt to access invalid positions on the board, making it a robust solution for all possible character transitions.