Count of Integers
<p>You are given two numeric strings <code>num1</code> and <code>num2</code> and two integers <code>max_sum</code> and <code>min_sum</code>. We denote an integer <code>x</code> to be <em>good</em> if:</p>

<ul>
	<li><code>num1 &lt;= x &lt;= num2</code></li>
	<li><code>min_sum &lt;= digit_sum(x) &lt;= max_sum</code>.</li>
</ul>

<p>Return <em>the number of good integers</em>. Since the answer may be large, return it modulo <code>10<sup>9</sup> + 7</code>.</p>

<p>Note that <code>digit_sum(x)</code> denotes the sum of the digits of <code>x</code>.</p>

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

<pre>
<strong>Input:</strong> num1 = &quot;1&quot;, num2 = &quot;12&quot;, <code>min_sum</code> = 1, max_sum = 8
<strong>Output:</strong> 11
<strong>Explanation:</strong> There are 11 integers whose sum of digits lies between 1 and 8 are 1,2,3,4,5,6,7,8,10,11, and 12. Thus, we return 11.
</pre>

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

<pre>
<strong>Input:</strong> num1 = &quot;1&quot;, num2 = &quot;5&quot;, <code>min_sum</code> = 1, max_sum = 5
<strong>Output:</strong> 5
<strong>Explanation:</strong> The 5 integers whose sum of digits lies between 1 and 5 are 1,2,3,4, and 5. Thus, we return 5.
</pre>

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

<ul>
	<li><code>1 &lt;= num1 &lt;= num2 &lt;= 10<sup>22</sup></code></li>
	<li><code>1 &lt;= min_sum &lt;= max_sum &lt;= 400</code></li>
</ul>
Math; String; Dynamic Programming
class Solution:
    def count(self, num1: str, num2: str, min_sum: int, max_sum: int) -> int:
        @cache
        def dfs(pos: int, s: int, limit: bool) -> int:
            if pos >= len(num):
                return int(min_sum <= s <= max_sum)
            up = int(num[pos]) if limit else 9
            return (
                sum(dfs(pos + 1, s + i, limit and i == up) for i in range(up + 1)) % mod
            )

        mod = 10**9 + 7
        num = num2
        a = dfs(0, 0, True)
        dfs.cache_clear()
        num = str(int(num1) - 1)
        b = dfs(0, 0, True)
        return (a - b) % mod
