<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>

class Solution:
    def minimumBeautifulSubstrings(self, s: str) -> int:
        powers = set(["1"])
        i = 1
        while 5 ** i <= 32767:
            powers.add(bin(5**i)[2:])
            i += 1
        #print(powers)

        found = [False for i in range(len(s))]
        found.append(1)
        for j in range(len(s)-1, -1, -1):
            for z in powers:
                #print(j, j+len(z), z, s[j:])
                if j + len(z) < len(s)+1 and found[j+len(z)] and s[j:j+len(z)] == z:#might need to be -1 index
                    #print('here', j, found[j+len(z)] + 1)
                    val2 = found[j] if found[j] else float("inf")
                    found[j] = min(found[j+len(z)] + 1, val2)
                else:
                    continue
            #print(found)
        return found[0]-1 if found[0] > 0 else -1
        
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