LeetCode Q 1138 - Alphabet Board Path

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.

Solution:

Be careful about the move sequence. 'U' should appear before R, and L should appear before D.

Code:

public String alphabetBoardPath(String target) {
  Map<Character, int[]> map = new HashMap<>(); 
  
  for (int i = 0; i < 26; i++) {
    map.put((char)('a' + i), new int[]{i / 5, i % 5});
  }

  StringBuilder sb = new StringBuilder();
  int r = 0, c = 0;
  for (char ch: target.toCharArray()) {
    int nr = map.get(ch)[0], nc = map.get(ch)[1];
    while (r > nr) { r--; sb.append('U')};
    while (c < nc) { c++; sb.append('R')};
    while (c > nc) { c--; sb.append('L')};
    while (r < nr) { r++; sb.append('D')};
    sb.append('!');
    r = nr; c = nc;
  }

  return sb.toString();
}

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