Count Ways To Build Good Strings
<p>Given the integers <code>zero</code>, <code>one</code>, <code>low</code>, and <code>high</code>, we can construct a string by starting with an empty string, and then at each step perform either of the following:</p>

<ul>
	<li>Append the character <code>&#39;0&#39;</code> <code>zero</code> times.</li>
	<li>Append the character <code>&#39;1&#39;</code> <code>one</code> times.</li>
</ul>

<p>This can be performed any number of times.</p>

<p>A <strong>good</strong> string is a string constructed by the above process having a <strong>length</strong> between <code>low</code> and <code>high</code> (<strong>inclusive</strong>).</p>

<p>Return <em>the number of <strong>different</strong> good strings that can be constructed satisfying these properties.</em> Since the answer can be 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> low = 3, high = 3, zero = 1, one = 1
<strong>Output:</strong> 8
<strong>Explanation:</strong> 
One possible valid good string is &quot;011&quot;. 
It can be constructed as follows: &quot;&quot; -&gt; &quot;0&quot; -&gt; &quot;01&quot; -&gt; &quot;011&quot;. 
All binary strings from &quot;000&quot; to &quot;111&quot; are good strings in this example.
</pre>

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

<pre>
<strong>Input:</strong> low = 2, high = 3, zero = 1, one = 2
<strong>Output:</strong> 5
<strong>Explanation:</strong> The good strings are &quot;00&quot;, &quot;11&quot;, &quot;000&quot;, &quot;110&quot;, and &quot;011&quot;.
</pre>

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

<ul>
	<li><code>1 &lt;= low&nbsp;&lt;= high&nbsp;&lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= zero, one &lt;= low</code></li>
</ul>
Dynamic Programming
func countGoodStrings(low int, high int, zero int, one int) int {
	f := make([]int, high+1)
	for i := range f {
		f[i] = -1
	}
	const mod int = 1e9 + 7
	var dfs func(i int) int
	dfs = func(i int) int {
		if i > high {
			return 0
		}
		if f[i] != -1 {
			return f[i]
		}
		ans := 0
		if i >= low && i <= high {
			ans++
		}
		ans += dfs(i+zero) + dfs(i+one)
		ans %= mod
		f[i] = ans
		return ans
	}
	return dfs(0)
}