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
import java.math.BigInteger;

class Solution {
    private final int mod = (int) 1e9 + 7;
    private Integer[][] f;
    private String num;
    private int min;
    private int max;

    public int count(String num1, String num2, int min_sum, int max_sum) {
        min = min_sum;
        max = max_sum;
        num = num2;
        f = new Integer[23][220];
        int a = dfs(0, 0, true);
        num = new BigInteger(num1).subtract(BigInteger.ONE).toString();
        f = new Integer[23][220];
        int b = dfs(0, 0, true);
        return (a - b + mod) % mod;
    }

    private int dfs(int pos, int s, boolean limit) {
        if (pos >= num.length()) {
            return s >= min && s <= max ? 1 : 0;
        }
        if (!limit && f[pos][s] != null) {
            return f[pos][s];
        }
        int ans = 0;
        int up = limit ? num.charAt(pos) - '0' : 9;
        for (int i = 0; i <= up; ++i) {
            ans = (ans + dfs(pos + 1, s + i, limit && i == up)) % mod;
        }
        if (!limit) {
            f[pos][s] = ans;
        }
        return ans;
    }
}