<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 {
public:
    string alphabetBoardPath(string target) {
        pair<int, int> currentPos = {0, 0};
        string path;
        int row, col;
        for (const char& c : target) {
            row = (c - 'a') / 5;
            col = (c - 'a') % 5;
            if (row < currentPos.first)
                path += string(currentPos.first - row, 'U');
            if (col < currentPos.second)
                path += string(currentPos.second - col, 'L');
            if (row > currentPos.first)
                path += string(row - currentPos.first, 'D');
            if (col > currentPos.second)
                path += string(col - currentPos.second, 'R');
            path += '!';
            currentPos.first = row;
            currentPos.second = col;
        }
        return path;
    }
};
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