Count the Number of Powerful Integers
<p>You are given three integers <code>start</code>, <code>finish</code>, and <code>limit</code>. You are also given a <strong>0-indexed</strong> string <code>s</code> representing a <strong>positive</strong> integer.</p>

<p>A <strong>positive</strong> integer <code>x</code> is called <strong>powerful</strong> if it ends with <code>s</code> (in other words, <code>s</code> is a <strong>suffix</strong> of <code>x</code>) and each digit in <code>x</code> is at most <code>limit</code>.</p>

<p>Return <em>the <strong>total</strong> number of powerful integers in the range</em> <code>[start..finish]</code>.</p>

<p>A string <code>x</code> is a suffix of a string <code>y</code> if and only if <code>x</code> is a substring of <code>y</code> that starts from some index (<strong>including </strong><code>0</code>) in <code>y</code> and extends to the index <code>y.length - 1</code>. For example, <code>25</code> is a suffix of <code>5125</code> whereas <code>512</code> is not.</p>

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

<pre>
<strong>Input:</strong> start = 1, finish = 6000, limit = 4, s = &quot;124&quot;
<strong>Output:</strong> 5
<strong>Explanation:</strong> The powerful integers in the range [1..6000] are 124, 1124, 2124, 3124, and, 4124. All these integers have each digit &lt;= 4, and &quot;124&quot; as a suffix. Note that 5124 is not a powerful integer because the first digit is 5 which is greater than 4.
It can be shown that there are only 5 powerful integers in this range.
</pre>

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

<pre>
<strong>Input:</strong> start = 15, finish = 215, limit = 6, s = &quot;10&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> The powerful integers in the range [15..215] are 110 and 210. All these integers have each digit &lt;= 6, and &quot;10&quot; as a suffix.
It can be shown that there are only 2 powerful integers in this range.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> start = 1000, finish = 2000, limit = 4, s = &quot;3000&quot;
<strong>Output:</strong> 0
<strong>Explanation:</strong> All integers in the range [1000..2000] are smaller than 3000, hence &quot;3000&quot; cannot be a suffix of any integer in this range.
</pre>

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

<ul>
	<li><code>1 &lt;= start &lt;= finish &lt;= 10<sup>15</sup></code></li>
	<li><code>1 &lt;= limit &lt;= 9</code></li>
	<li><code>1 &lt;= s.length &lt;= floor(log<sub>10</sub>(finish)) + 1</code></li>
	<li><code>s</code> only consists of numeric digits which are at most <code>limit</code>.</li>
	<li><code>s</code> does not have leading zeros.</li>
</ul>
Math; String; Dynamic Programming
class Solution {
    private String s;
    private String t;
    private Long[] f;
    private int limit;

    public long numberOfPowerfulInt(long start, long finish, int limit, String s) {
        this.s = s;
        this.limit = limit;
        t = String.valueOf(start - 1);
        f = new Long[20];
        long a = dfs(0, true);
        t = String.valueOf(finish);
        f = new Long[20];
        long b = dfs(0, true);
        return b - a;
    }

    private long dfs(int pos, boolean lim) {
        if (t.length() < s.length()) {
            return 0;
        }
        if (!lim && f[pos] != null) {
            return f[pos];
        }
        if (t.length() - pos == s.length()) {
            return lim ? (s.compareTo(t.substring(pos)) <= 0 ? 1 : 0) : 1;
        }
        int up = lim ? t.charAt(pos) - '0' : 9;
        up = Math.min(up, limit);
        long ans = 0;
        for (int i = 0; i <= up; ++i) {
            ans += dfs(pos + 1, lim && i == (t.charAt(pos) - '0'));
        }
        if (!lim) {
            f[pos] = ans;
        }
        return ans;
    }
}