<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:
    pair<int, int> findPos(char c){
        int row = (c - 'a')/5;
        int col = (c - 'a')%5;
        return {row, col};
    }

    string alphabetBoardPath(string target) {
        string ans = "";
        int currRow = 0, currCol = 0;
        for(auto c : target){
            pair<int, int> curr = findPos(c);
            int r = curr.first;
            int col = curr.second;
            int rdiff = r - currRow;
            int cdiff = col - currCol;
            string t1 = "", t2 = "";
            if(cdiff > 0){
                for(int i = 0; i < cdiff; i++){
                    t1 += "R";
                }
            }
            else{
                for(int i = 0; i < -cdiff; i++){
                    t1 += "L";
                }
            }
            if(rdiff > 0){
                for(int i = 0; i < rdiff; i++){
                    t2 += "D";
                }
            }
            else{
                for(int i = 0; i < -rdiff; i++){
                    t2 += "U";
                }
            }
            if(currRow == 5){
                ans += t2 + t1;
            }
            else{
                ans += t1 + t2;
            }
            ans += "!";
            currRow = r;
            currCol = col;
        }

        return ans;
    }
};
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