LeetCode #1138 — MEDIUM

Alphabet Board Path

Move from brute-force thinking to an efficient approach using hash map strategy.

Solve on LeetCode

The Problem

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map

Example 1

"leet"

Example 2

"code"

Step 02

Core Insight

What unlocks the optimal approach

Create a hashmap from letter to position on the board.
Now for each letter, try moving there in steps, where at each step you check if it is inside the boundaries of the board.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1138: Alphabet Board Path
class Solution {
    public String alphabetBoardPath(String target) {
        StringBuilder ans = new StringBuilder();
        int i = 0, j = 0;
        for (int k = 0; k < target.length(); ++k) {
            int v = target.charAt(k) - 'a';
            int x = v / 5, y = v % 5;
            while (j > y) {
                --j;
                ans.append('L');
            }
            while (i > x) {
                --i;
                ans.append('U');
            }
            while (j < y) {
                ++j;
                ans.append('R');
            }
            while (i < x) {
                ++i;
                ans.append('D');
            }
            ans.append("!");
        }
        return ans.toString();
    }
}

// Accepted solution for LeetCode #1138: Alphabet Board Path
func alphabetBoardPath(target string) string {
	ans := []byte{}
	var i, j int
	for _, c := range target {
		v := int(c - 'a')
		x, y := v/5, v%5
		for j > y {
			j--
			ans = append(ans, 'L')
		}
		for i > x {
			i--
			ans = append(ans, 'U')
		}
		for j < y {
			j++
			ans = append(ans, 'R')
		}
		for i < x {
			i++
			ans = append(ans, 'D')
		}
		ans = append(ans, '!')
	}
	return string(ans)
}

# Accepted solution for LeetCode #1138: Alphabet Board Path
class Solution:
    def alphabetBoardPath(self, target: str) -> str:
        i = j = 0
        ans = []
        for c in target:
            v = ord(c) - ord("a")
            x, y = v // 5, v % 5
            while j > y:
                j -= 1
                ans.append("L")
            while i > x:
                i -= 1
                ans.append("U")
            while j < y:
                j += 1
                ans.append("R")
            while i < x:
                i += 1
                ans.append("D")
            ans.append("!")
        return "".join(ans)

// Accepted solution for LeetCode #1138: Alphabet Board Path
struct Solution;

impl Solution {
    fn alphabet_board_path(target: String) -> String {
        let mut pos: Vec<(i32, i32)> = vec![];
        for i in 0..5 {
            for j in 0..5 {
                pos.push((i, j));
            }
        }
        pos.push((5, 0));

        let mut v = vec!['a'];
        for c in target.chars() {
            v.push(c);
        }
        let n = v.len();
        let mut res = "".to_string();
        for i in 1..n {
            let curr = pos[(v[i] as u8 - b'a') as usize];
            let prev = pos[(v[i - 1] as u8 - b'a') as usize];
            let mut r = curr.0 - prev.0;
            let mut c = curr.1 - prev.1;
            while r < 0 {
                res.push('U');
                r += 1;
            }
            while c < 0 {
                res.push('L');
                c += 1;
            }
            while r > 0 {
                res.push('D');
                r -= 1;
            }
            while c > 0 {
                res.push('R');
                c -= 1;
            }
            res.push('!');
        }
        res
    }
}

#[test]
fn test() {
    let target = "leet".to_string();
    let res = "DDR!UURRR!!DDD!".to_string();
    assert_eq!(Solution::alphabet_board_path(target), res);
    let target = "code".to_string();
    let res = "RR!DDRR!UUL!R!".to_string();
    assert_eq!(Solution::alphabet_board_path(target), res);
}

// Accepted solution for LeetCode #1138: Alphabet Board Path
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1138: Alphabet Board Path
// class Solution {
//     public String alphabetBoardPath(String target) {
//         StringBuilder ans = new StringBuilder();
//         int i = 0, j = 0;
//         for (int k = 0; k < target.length(); ++k) {
//             int v = target.charAt(k) - 'a';
//             int x = v / 5, y = v % 5;
//             while (j > y) {
//                 --j;
//                 ans.append('L');
//             }
//             while (i > x) {
//                 --i;
//                 ans.append('U');
//             }
//             while (j < y) {
//                 ++j;
//                 ans.append('R');
//             }
//             while (i < x) {
//                 ++i;
//                 ans.append('D');
//             }
//             ans.append("!");
//         }
//         return ans.toString();
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP

O(n) time

O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.