<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>

MOD = 10**9 + 7
class Solution:
    def countGoodStrings(self, low: int, high: int, zero: int, one: int) -> int:
        def get_dp_array(length, zero, one):
            dp_arr = [0] * (length + 1)
            dp_arr[0] = 1
            for index in range(1, length + 1):
                left = dp_arr[index - one] if index - one >= 0 else 0
                right = dp_arr[index - zero] if index - zero >= 0 else 0
                dp_arr[index] = (left + right + 1)
            return dp_arr

        dp_array = get_dp_array(high, zero, one)
        return (dp_array[high] - dp_array[low - 1]) % MOD

543