<p>On an alphabet board, we start at position <code>(0, 0)</code>, corresponding to character&nbsp;<code>board[0][0]</code>.</p>

<p>Here, <code>board = [&quot;abcde&quot;, &quot;fghij&quot;, &quot;klmno&quot;, &quot;pqrst&quot;, &quot;uvwxy&quot;, &quot;z&quot;]</code>, as shown in the diagram below.</p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2019/07/28/azboard.png" style="width: 250px; height: 317px;" /></p>

<p>We may make the following moves:</p>

<ul>
	<li><code>&#39;U&#39;</code> moves our position up one row, if the position exists on the board;</li>
	<li><code>&#39;D&#39;</code> moves our position down one row, if the position exists on the board;</li>
	<li><code>&#39;L&#39;</code> moves our position left one column, if the position exists on the board;</li>
	<li><code>&#39;R&#39;</code> moves our position right one column, if the position exists on the board;</li>
	<li><code>&#39;!&#39;</code>&nbsp;adds the character <code>board[r][c]</code> at our current position <code>(r, c)</code>&nbsp;to the&nbsp;answer.</li>
</ul>

<p>(Here, the only positions that exist on the board are positions with letters on them.)</p>

<p>Return a sequence of moves that makes our answer equal to <code>target</code>&nbsp;in the minimum number of moves.&nbsp; You may return any path that does so.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> target = "leet"
<strong>Output:</strong> "DDR!UURRR!!DDD!"
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> target = "code"
<strong>Output:</strong> "RR!DDRR!UUL!R!"
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= target.length &lt;= 100</code></li>
	<li><code>target</code> consists only of English lowercase letters.</li>
</ul>
class Solution:
    MAPPING = {
        ltr: (i // 5, i % 5)
        for i, ltr in enumerate(ascii_lowercase)
    }
    def alphabetBoardPath(self, target: str) -> str:
        x, y = 0, 0
        path = ''

        for c in target:
            n_x, n_y = self.MAPPING[c]

            if n_y < y:
                path += 'L' * (y - n_y)
            if n_x < x:
                path += 'U' * (x - n_x)
            if n_x > x:
                path += 'D' * (n_x - x)
            if n_y > y:
                path += 'R' * (n_y - y)
            
            path += '!'
            x, y = n_x, n_y
        
        return path
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