How do you solve Alphabet Board Path on LeetCode?

Alphabet Board Path (LeetCode #1138) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Mapping characters to their positions helps in quick access.

What is the brute force solution for Alphabet Board Path?

The brute force approach for Alphabet Board Path is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves calculating the position of each character in the target string on the board and moving step by step to reach each character from the current position. This method is straightforward but inefficient as it checks each move individually. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Alphabet Board Path?

Alphabet Board Path uses the following concepts: Hash Table, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Alphabet Board Path?

Always clarify the problem and constraints before jumping into coding. Think about edge cases and how they might affect your solution. Practice explaining your thought process as you code; it helps in interviews.

What are common mistakes when solving Alphabet Board Path?

Not considering edge cases like moving to 'z' which requires special handling. Overcomplicating the movement logic instead of simplifying it with coordinate differences.

#1138

Alphabet Board Path

Medium

Hash TableStringHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution improves efficiency by reducing unnecessary moves. Instead of moving in a straight line to each character, we can calculate the direction to move in a single pass, minimizing the number of moves.

⚙️

Algorithm

3 steps

1Step 1: Create a mapping of each character to its position on the board.
2Step 2: For each character in the target string, determine the target position and calculate the difference in rows and columns from the current position.
3Step 3: Move vertically and horizontally based on the calculated differences, appending the necessary moves to the result string.

solution.py20 lines

1# Full working Python code
2class Solution:
3    def alphabetBoardPath(self, target: str) -> str:
4        board = ["abcde", "fghij", "klmno", "pqrst", "uvwxy", "z"]
5        pos = {ch: (i, j) for i, row in enumerate(board) for j, ch in enumerate(row)}
6        result = []
7        x, y = 0, 0
8        for char in target:
9            target_x, target_y = pos[char]
10            if target_x < x:
11                result.append('U' * (x - target_x))
12            if target_x > x:
13                result.append('D' * (target_x - x))
14            if target_y < y:
15                result.append('L' * (y - target_y))
16            if target_y > y:
17                result.append('R' * (target_y - y))
18            result.append('!')
19            x, y = target_x, target_y
20        return ''.join(result)

ℹ

Complexity note: The time complexity is O(n) because we process each character in the target string once, and the space complexity is O(n) due to the storage of character positions in the hashmap.

1Mapping characters to their positions helps in quick access.
2Minimizing moves by calculating differences in coordinates reduces unnecessary steps.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.