<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 ans;
    void fn(string s,int ind){
        int c1=s[ind]-'a',c2=s[ind+1]-'a';
        if(c1==c2) return;
        int x1=c1/5,y1=c1%5,x2=c2/5,y2=c2%5;
        if(s[ind+1]=='z'){
            ans+=string(y1,'L');
            ans+=string(5-x1,'D');
            return;
        }
        if(x1<x2) ans+=string(abs(x1-x2),'D');
        else ans+=string(abs(x1-x2),'U');
        if(y1<y2) ans+=string(abs(y1-y2),'R');
        else ans+=string(abs(y1-y2),'L');
    }
    string alphabetBoardPath(string s) {
        int n=s.size();
        s='a'+s;
        for(int j=0;j<n;j++) {fn(s,j); ans+='!';}
        return ans;
    }
};
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