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Alphabet Board Path

Medium

Asked by:

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Topics:

Strings

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.

Try coding it in LeetCode

Alphabet Board Path

Medium

Asked by:

9 views

Topics:

Strings

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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the target string contain characters not present on the alphabet board (e.g., special characters or uppercase letters)?
If the target string is empty, what should the output be?
Is the target string guaranteed to be valid (e.g., only lowercase English letters)?
Is there a preferred direction to move when multiple paths exist to the same character (e.g., should I prioritize up/down over left/right)?
Should the path be optimized for the fewest number of moves, or is any valid path acceptable?

Brute Force Solution

Approach

The brute force approach to finding the shortest path on the alphabet board is to explore every possible sequence of moves. We try every combination of up, down, left, and right movements until we find a path that spells out the target word. Since that could take a very long time, we will need a better approach in a real interview, but this is the basic idea.

Here's how the algorithm would work step-by-step:

Start at the top-left letter of the board (the letter 'a').
For the first letter of the target word, try moving up, down, left, and right one step at a time.
If a move gets you closer to the first letter of the word, keep going in that direction. If not, try a different direction.
Keep track of the path you take to get to that first letter.
Now, from that first letter, repeat the process for the second letter of the target word: try all the movements (up, down, left, right).
Again, keep track of the path to the second letter.
Continue this process for every letter in the target word, always recording the path taken.
Because this is brute force, if you find a path that spells the whole word, just continue trying different paths, just in case there's an even shorter one.
After exploring many, many different possible paths, compare the lengths of all the paths that successfully spelled the word.
Finally, choose the shortest path among all the successful paths as the answer.

Code Implementation

def alphabet_board_path_brute_force(target):
    board = ["abcde", "fghij", "klmno", "pqrst", "uvwxyz"]

    def find_char(character):
        for row_index, row in enumerate(board):
            if character in row:
                col_index = row.index(character)
                return row_index, col_index
        return None

    def get_path(start_row, start_col, end_row, end_col):
        path = ""
        
        # Move up if needed before left
        while start_row > end_row:
            path += "U"
            start_row -= 1
            
        # Move left if needed
        while start_col > end_col:
            path += "L"
            start_col -= 1

        # Move right if needed
        while start_col < end_col:
            path += "R"
            start_col += 1
            
        # Move down if needed last
        while start_row < end_row:
            path += "D"
            start_row += 1

        path += "!"
        return path

    current_row, current_col = 0, 0
    total_path = ""
    
    # Iterate over each letter
    for character in target:
        target_row, target_col = find_char(character)

        # Construct the path to each character
        path_to_character = get_path(current_row, current_col, target_row, target_col)
        total_path += path_to_character
        
        current_row, current_col = target_row, target_col

    return total_path

Big(O) Analysis

Time Complexity

O(4^m * n) – The described brute force approach explores all possible paths on the alphabet board. For each letter in the target word (of length n), we explore up to four possible moves (up, down, left, right). This leads to a branching factor of 4 for each position in the target word. In the worst case, we need to explore paths of maximum length m (related to the total number of moves to form the word, and depends on the arrangement of letters on the board), giving us a search space of approximately 4^m. And since we need to continue after finding a valid path to ensure there is no shorter path, the total number of operations depends on the length of the target word (n). Thus, the time complexity is roughly O(4^m * n), where 'm' represents the maximum number of moves possible to reach the farthest letter, and 'n' is the length of the target word.

Space Complexity

O(N!) – The brute force approach explores all possible paths of movements (up, down, left, right) until a target word of length N is spelled out. The space used to store each possible path grows factorially, as the algorithm must keep track of each move for potentially every single letter permutation. While not explicitly stated which datastructure holds all the paths, the need to compare all paths implies that all N! combinations are being held in memory. Therefore, the space complexity is O(N!).

Optimal Solution

Approach

The optimal approach involves navigating a 5x5 alphabet board to spell a target word. We will determine the sequence of moves required to travel from 'a' to each letter in the target, optimizing for efficiency. The core idea is to use the shortest path using vertical and horizontal movements, and handle 'z' specially.

Here's how the algorithm would work step-by-step:

Begin at the starting position 'a'.
For each letter in the target word, calculate the horizontal and vertical distance from your current position on the board to the target letter's position.
Prioritize moving vertically ('U' or 'D') before moving horizontally ('L' or 'R'), except when moving to 'z'. Moving to 'z' requires moving horizontally first.
If moving to 'z', move left ('L') to ensure there is space to move down ('D') to reach 'z'. Otherwise, when moving away from 'z', moving vertically before horizontally generally minimizes unnecessary movements around 'z'.
Append the appropriate move character ('U', 'D', 'L', 'R') to your path string based on the calculated distances. Make sure that you are only adding the appropriate characters to the path. For instance, only 'U' if vertical distance is negative.
After reaching each letter, add a '!' to your path string to indicate selecting that letter.
Update your current position to be the position of the letter you just reached.
Repeat steps 2-7 for each letter in the target word.
Return the completed path string, which represents the sequence of moves needed to spell the target word.

Code Implementation

def alphabet_board_path(target_word):
    current_row = 0
    current_column = 0
    path = ""

    for character in target_word:
        target_row = (ord(character) - ord('a')) // 5
        target_column = (ord(character) - ord('a')) % 5

        # Prioritize left movement if target is 'z'.
        if character == 'z':
            while current_column > target_column:
                path += 'L'
                current_column -= 1
        
        # Move vertically first, except when target is 'z'.
        while current_row > target_row:
            path += 'U'
            current_row -= 1

        while current_row < target_row:
            path += 'D'
            current_row += 1

        # Move horizontally to reach target.
        while current_column < target_column:
            path += 'R'
            current_column += 1

        while current_column > target_column:
            path += 'L'
            current_column -= 1

        path += '!'

        # Update current position for the next character.
        current_row = target_row
        current_column = target_column

    return path

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through each character in the target word once. The cost associated with each character involves calculating the vertical and horizontal distance to move from the current position to the target character's position on the alphabet board and then constructing the path. The distance calculations and path construction involve a fixed number of operations per character, independent of the length of the target word. Thus, if n is the length of the target word, the overall time complexity is O(n).

Space Complexity

O(N) – The algorithm constructs a path string to store the sequence of moves. The length of this string is proportional to the number of characters in the target word, as each character requires a series of moves. Therefore, the space required to store the path string grows linearly with the length of the target word, which we can denote as N. Thus the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Null or empty target string	Return an empty string since there is no path to generate for an empty target.
Target string contains characters not on the alphabet board	Throw an IllegalArgumentException or return an error string since the character is not reachable.
Target string contains only 'a' characters	Return a string of '!' characters with the same length as the input since it starts at 'a'.
Target string contains only 'z' characters	The path to 'z' must be handled correctly with 'D' and 'L' moves before appending '!'.
Very long target string	The solution should scale linearly with the length of the target string in terms of time complexity.
Target requires zigzagging paths (e.g., 'abc')	The movement calculation must correctly handle both row and column changes, and account for 'z' at (4,0).
Moving from 'z' to another character	The movement from 'z' should always prioritize moving up before moving right.
Target string starts with 'z'	The solution should be able to correctly calculate the path starting from (0,0) to 'z' (4,0).

Null or empty target string

How to Handle:

Return an empty string since there is no path to generate for an empty target.

Target string contains characters not on the alphabet board

How to Handle:

Throw an IllegalArgumentException or return an error string since the character is not reachable.

Target string contains only 'a' characters

How to Handle:

Return a string of '!' characters with the same length as the input since it starts at 'a'.

Target string contains only 'z' characters

How to Handle:

The path to 'z' must be handled correctly with 'D' and 'L' moves before appending '!'.

Very long target string

How to Handle:

The solution should scale linearly with the length of the target string in terms of time complexity.

Target requires zigzagging paths (e.g., 'abc')

How to Handle:

The movement calculation must correctly handle both row and column changes, and account for 'z' at (4,0).

Moving from 'z' to another character

How to Handle:

The movement from 'z' should always prioritize moving up before moving right.

Target string starts with 'z'

How to Handle:

The solution should be able to correctly calculate the path starting from (0,0) to 'z' (4,0).

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the target string contain characters not present on the alphabet board (e.g., special characters or uppercase letters)?
If the target string is empty, what should the output be?
Is the target string guaranteed to be valid (e.g., only lowercase English letters)?
Is there a preferred direction to move when multiple paths exist to the same character (e.g., should I prioritize up/down over left/right)?
Should the path be optimized for the fewest number of moves, or is any valid path acceptable?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start at the top-left letter of the board (the letter 'a').
For the first letter of the target word, try moving up, down, left, and right one step at a time.
If a move gets you closer to the first letter of the word, keep going in that direction. If not, try a different direction.
Keep track of the path you take to get to that first letter.
Now, from that first letter, repeat the process for the second letter of the target word: try all the movements (up, down, left, right).
Again, keep track of the path to the second letter.
Continue this process for every letter in the target word, always recording the path taken.
Because this is brute force, if you find a path that spells the whole word, just continue trying different paths, just in case there's an even shorter one.
After exploring many, many different possible paths, compare the lengths of all the paths that successfully spelled the word.
Finally, choose the shortest path among all the successful paths as the answer.

Code Implementation

def alphabet_board_path_brute_force(target):
    board = ["abcde", "fghij", "klmno", "pqrst", "uvwxyz"]

    def find_char(character):
        for row_index, row in enumerate(board):
            if character in row:
                col_index = row.index(character)
                return row_index, col_index
        return None

    def get_path(start_row, start_col, end_row, end_col):
        path = ""
        
        # Move up if needed before left
        while start_row > end_row:
            path += "U"
            start_row -= 1
            
        # Move left if needed
        while start_col > end_col:
            path += "L"
            start_col -= 1

        # Move right if needed
        while start_col < end_col:
            path += "R"
            start_col += 1
            
        # Move down if needed last
        while start_row < end_row:
            path += "D"
            start_row += 1

        path += "!"
        return path

    current_row, current_col = 0, 0
    total_path = ""
    
    # Iterate over each letter
    for character in target:
        target_row, target_col = find_char(character)

        # Construct the path to each character
        path_to_character = get_path(current_row, current_col, target_row, target_col)
        total_path += path_to_character
        
        current_row, current_col = target_row, target_col

    return total_path

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Begin at the starting position 'a'.
For each letter in the target word, calculate the horizontal and vertical distance from your current position on the board to the target letter's position.
Prioritize moving vertically ('U' or 'D') before moving horizontally ('L' or 'R'), except when moving to 'z'. Moving to 'z' requires moving horizontally first.
If moving to 'z', move left ('L') to ensure there is space to move down ('D') to reach 'z'. Otherwise, when moving away from 'z', moving vertically before horizontally generally minimizes unnecessary movements around 'z'.
Append the appropriate move character ('U', 'D', 'L', 'R') to your path string based on the calculated distances. Make sure that you are only adding the appropriate characters to the path. For instance, only 'U' if vertical distance is negative.
After reaching each letter, add a '!' to your path string to indicate selecting that letter.
Update your current position to be the position of the letter you just reached.
Repeat steps 2-7 for each letter in the target word.
Return the completed path string, which represents the sequence of moves needed to spell the target word.

Code Implementation

def alphabet_board_path(target_word):
    current_row = 0
    current_column = 0
    path = ""

    for character in target_word:
        target_row = (ord(character) - ord('a')) // 5
        target_column = (ord(character) - ord('a')) % 5

        # Prioritize left movement if target is 'z'.
        if character == 'z':
            while current_column > target_column:
                path += 'L'
                current_column -= 1
        
        # Move vertically first, except when target is 'z'.
        while current_row > target_row:
            path += 'U'
            current_row -= 1

        while current_row < target_row:
            path += 'D'
            current_row += 1

        # Move horizontally to reach target.
        while current_column < target_column:
            path += 'R'
            current_column += 1

        while current_column > target_column:
            path += 'L'
            current_column -= 1

        path += '!'

        # Update current position for the next character.
        current_row = target_row
        current_column = target_column

    return path

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty target string	Return an empty string since there is no path to generate for an empty target.
Target string contains characters not on the alphabet board	Throw an IllegalArgumentException or return an error string since the character is not reachable.
Target string contains only 'a' characters	Return a string of '!' characters with the same length as the input since it starts at 'a'.
Target string contains only 'z' characters	The path to 'z' must be handled correctly with 'D' and 'L' moves before appending '!'.
Very long target string	The solution should scale linearly with the length of the target string in terms of time complexity.
Target requires zigzagging paths (e.g., 'abc')	The movement calculation must correctly handle both row and column changes, and account for 'z' at (4,0).
Moving from 'z' to another character	The movement from 'z' should always prioritize moving up before moving right.
Target string starts with 'z'	The solution should be able to correctly calculate the path starting from (0,0) to 'z' (4,0).

Null or empty target string

How to Handle:

Return an empty string since there is no path to generate for an empty target.

Target string contains characters not on the alphabet board

How to Handle:

Throw an IllegalArgumentException or return an error string since the character is not reachable.

Target string contains only 'a' characters

How to Handle:

Return a string of '!' characters with the same length as the input since it starts at 'a'.

Target string contains only 'z' characters

How to Handle:

The path to 'z' must be handled correctly with 'D' and 'L' moves before appending '!'.

Very long target string

How to Handle:

The solution should scale linearly with the length of the target string in terms of time complexity.

Target requires zigzagging paths (e.g., 'abc')

How to Handle:

The movement calculation must correctly handle both row and column changes, and account for 'z' at (4,0).

Moving from 'z' to another character

How to Handle:

The movement from 'z' should always prioritize moving up before moving right.

Target string starts with 'z'

How to Handle:

The solution should be able to correctly calculate the path starting from (0,0) to 'z' (4,0).