Number of Ways to Form a Target String Given a Dictionary
<p>You are given a list of strings of the <strong>same length</strong> <code>words</code> and a string <code>target</code>.</p>

<p>Your task is to form <code>target</code> using the given <code>words</code> under the following rules:</p>

<ul>
	<li><code>target</code> should be formed from left to right.</li>
	<li>To form the <code>i<sup>th</sup></code> character (<strong>0-indexed</strong>) of <code>target</code>, you can choose the <code>k<sup>th</sup></code> character of the <code>j<sup>th</sup></code> string in <code>words</code> if <code>target[i] = words[j][k]</code>.</li>
	<li>Once you use the <code>k<sup>th</sup></code> character of the <code>j<sup>th</sup></code> string of <code>words</code>, you <strong>can no longer</strong> use the <code>x<sup>th</sup></code> character of any string in <code>words</code> where <code>x &lt;= k</code>. In other words, all characters to the left of or at index <code>k</code> become unusuable for every string.</li>
	<li>Repeat the process until you form the string <code>target</code>.</li>
</ul>

<p><strong>Notice</strong> that you can use <strong>multiple characters</strong> from the <strong>same string</strong> in <code>words</code> provided the conditions above are met.</p>

<p>Return <em>the number of ways to form <code>target</code> from <code>words</code></em>. Since the answer may be too large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;acca&quot;,&quot;bbbb&quot;,&quot;caca&quot;], target = &quot;aba&quot;
<strong>Output:</strong> 6
<strong>Explanation:</strong> There are 6 ways to form target.
&quot;aba&quot; -&gt; index 0 (&quot;<u>a</u>cca&quot;), index 1 (&quot;b<u>b</u>bb&quot;), index 3 (&quot;cac<u>a</u>&quot;)
&quot;aba&quot; -&gt; index 0 (&quot;<u>a</u>cca&quot;), index 2 (&quot;bb<u>b</u>b&quot;), index 3 (&quot;cac<u>a</u>&quot;)
&quot;aba&quot; -&gt; index 0 (&quot;<u>a</u>cca&quot;), index 1 (&quot;b<u>b</u>bb&quot;), index 3 (&quot;acc<u>a</u>&quot;)
&quot;aba&quot; -&gt; index 0 (&quot;<u>a</u>cca&quot;), index 2 (&quot;bb<u>b</u>b&quot;), index 3 (&quot;acc<u>a</u>&quot;)
&quot;aba&quot; -&gt; index 1 (&quot;c<u>a</u>ca&quot;), index 2 (&quot;bb<u>b</u>b&quot;), index 3 (&quot;acc<u>a</u>&quot;)
&quot;aba&quot; -&gt; index 1 (&quot;c<u>a</u>ca&quot;), index 2 (&quot;bb<u>b</u>b&quot;), index 3 (&quot;cac<u>a</u>&quot;)
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;abba&quot;,&quot;baab&quot;], target = &quot;bab&quot;
<strong>Output:</strong> 4
<strong>Explanation:</strong> There are 4 ways to form target.
&quot;bab&quot; -&gt; index 0 (&quot;<u>b</u>aab&quot;), index 1 (&quot;b<u>a</u>ab&quot;), index 2 (&quot;ab<u>b</u>a&quot;)
&quot;bab&quot; -&gt; index 0 (&quot;<u>b</u>aab&quot;), index 1 (&quot;b<u>a</u>ab&quot;), index 3 (&quot;baa<u>b</u>&quot;)
&quot;bab&quot; -&gt; index 0 (&quot;<u>b</u>aab&quot;), index 2 (&quot;ba<u>a</u>b&quot;), index 3 (&quot;baa<u>b</u>&quot;)
&quot;bab&quot; -&gt; index 1 (&quot;a<u>b</u>ba&quot;), index 2 (&quot;ba<u>a</u>b&quot;), index 3 (&quot;baa<u>b</u>&quot;)
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= words.length &lt;= 1000</code></li>
	<li><code>1 &lt;= words[i].length &lt;= 1000</code></li>
	<li>All strings in <code>words</code> have the same length.</li>
	<li><code>1 &lt;= target.length &lt;= 1000</code></li>
	<li><code>words[i]</code> and <code>target</code> contain only lowercase English letters.</li>
</ul>
Array; String; Dynamic Programming
func numWays(words []string, target string) int {
	m, n := len(target), len(words[0])
	f := make([][]int, m)
	cnt := make([][26]int, n)
	for _, w := range words {
		for j, c := range w {
			cnt[j][c-'a']++
		}
	}
	for i := range f {
		f[i] = make([]int, n)
		for j := range f[i] {
			f[i][j] = -1
		}
	}
	const mod = 1e9 + 7
	var dfs func(i, j int) int
	dfs = func(i, j int) int {
		if i >= m {
			return 1
		}
		if j >= n {
			return 0
		}
		if f[i][j] != -1 {
			return f[i][j]
		}
		ans := dfs(i, j+1)
		ans = (ans + dfs(i+1, j+1)*cnt[j][target[i]-'a']) % mod
		f[i][j] = ans
		return ans
	}
	return dfs(0, 0)
}