Partition String Into Minimum Beautiful Substrings
<p>Given a binary string <code>s</code>, partition the string into one or more <strong>substrings</strong> such that each substring is <strong>beautiful</strong>.</p>

<p>A string is <strong>beautiful</strong> if:</p>

<ul>
	<li>It doesn&#39;t contain leading zeros.</li>
	<li>It&#39;s the <strong>binary</strong> representation of a number that is a power of <code>5</code>.</li>
</ul>

<p>Return <em>the <strong>minimum</strong> number of substrings in such partition. </em>If it is impossible to partition the string <code>s</code> into beautiful substrings,&nbsp;return <code>-1</code>.</p>

<p>A <strong>substring</strong> is a contiguous sequence of characters in a string.</p>

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

<pre>
<strong>Input:</strong> s = &quot;1011&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> We can paritition the given string into [&quot;101&quot;, &quot;1&quot;].
- The string &quot;101&quot; does not contain leading zeros and is the binary representation of integer 5<sup>1</sup> = 5.
- The string &quot;1&quot; does not contain leading zeros and is the binary representation of integer 5<sup>0</sup> = 1.
It can be shown that 2 is the minimum number of beautiful substrings that s can be partitioned into.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;111&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> We can paritition the given string into [&quot;1&quot;, &quot;1&quot;, &quot;1&quot;].
- The string &quot;1&quot; does not contain leading zeros and is the binary representation of integer 5<sup>0</sup> = 1.
It can be shown that 3 is the minimum number of beautiful substrings that s can be partitioned into.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;0&quot;
<strong>Output:</strong> -1
<strong>Explanation:</strong> We can not partition the given string into beautiful substrings.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 15</code></li>
	<li><code>s[i]</code> is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
</ul>
Hash Table; String; Dynamic Programming; Backtracking
class Solution {
    private Integer[] f;
    private String s;
    private Set<Long> ss = new HashSet<>();
    private int n;

    public int minimumBeautifulSubstrings(String s) {
        n = s.length();
        this.s = s;
        f = new Integer[n];
        long x = 1;
        for (int i = 0; i <= n; ++i) {
            ss.add(x);
            x *= 5;
        }
        int ans = dfs(0);
        return ans > n ? -1 : ans;
    }

    private int dfs(int i) {
        if (i >= n) {
            return 0;
        }
        if (s.charAt(i) == '0') {
            return n + 1;
        }
        if (f[i] != null) {
            return f[i];
        }
        long x = 0;
        int ans = n + 1;
        for (int j = i; j < n; ++j) {
            x = x << 1 | (s.charAt(j) - '0');
            if (ss.contains(x)) {
                ans = Math.min(ans, 1 + dfs(j + 1));
            }
        }
        return f[i] = ans;
    }
}