<p>You are given a <strong>0-indexed</strong> string <code>num</code> representing a non-negative integer.</p>

<p>In one operation, you can pick any digit of <code>num</code> and delete it. Note that if you delete all the digits of <code>num</code>, <code>num</code> becomes <code>0</code>.</p>

<p>Return <em>the <strong>minimum number of operations</strong> required to make</em> <code>num</code> <i>special</i>.</p>

<p>An integer <code>x</code> is considered <strong>special</strong> if it is divisible by <code>25</code>.</p>

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

<pre>
<strong>Input:</strong> num = &quot;2245047&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> Delete digits num[5] and num[6]. The resulting number is &quot;22450&quot; which is special since it is divisible by 25.
It can be shown that 2 is the minimum number of operations required to get a special number.</pre>

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

<pre>
<strong>Input:</strong> num = &quot;2908305&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> Delete digits num[3], num[4], and num[6]. The resulting number is &quot;2900&quot; which is special since it is divisible by 25.
It can be shown that 3 is the minimum number of operations required to get a special number.</pre>

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

<pre>
<strong>Input:</strong> num = &quot;10&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> Delete digit num[0]. The resulting number is &quot;0&quot; which is special since it is divisible by 25.
It can be shown that 1 is the minimum number of operations required to get a special number.

</pre>

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

<ul>
	<li><code>1 &lt;= num.length &lt;= 100</code></li>
	<li><code>num</code> only consists of digits <code>&#39;0&#39;</code> through <code>&#39;9&#39;</code>.</li>
	<li><code>num</code> does not contain any leading zeros.</li>
</ul>

class Solution:
    def minimumOperations(self, num: str) -> int:
        n = len(num)
        idx = -1
        for s in ['00','25','50','75']:
            i = n-1
            j = 1
            while i >= 0:
                if num[i] == s[j]:
                    j -= 1
                if j < 0:
                    idx = max(idx, i)
                    break
                i -= 1
            #print(idx)
        
        ans =  n-idx-2 if idx >= 0 else n
        if '0' in num:
            ans = min(ans, n-1)
        return ans
16500