Next Greater Numerically Balanced Number
<p>An integer <code>x</code> is <strong>numerically balanced</strong> if for every digit <code>d</code> in the number <code>x</code>, there are <strong>exactly</strong> <code>d</code> occurrences of that digit in <code>x</code>.</p>

<p>Given an integer <code>n</code>, return <em>the <strong>smallest numerically balanced</strong> number <strong>strictly greater</strong> than </em><code>n</code><em>.</em></p>

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

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 22
<strong>Explanation:</strong> 
22 is numerically balanced since:
- The digit 2 occurs 2 times. 
It is also the smallest numerically balanced number strictly greater than 1.
</pre>

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

<pre>
<strong>Input:</strong> n = 1000
<strong>Output:</strong> 1333
<strong>Explanation:</strong> 
1333 is numerically balanced since:
- The digit 1 occurs 1 time.
- The digit 3 occurs 3 times. 
It is also the smallest numerically balanced number strictly greater than 1000.
Note that 1022 cannot be the answer because 0 appeared more than 0 times.
</pre>

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

<pre>
<strong>Input:</strong> n = 3000
<strong>Output:</strong> 3133
<strong>Explanation:</strong> 
3133 is numerically balanced since:
- The digit 1 occurs 1 time.
- The digit 3 occurs 3 times.
It is also the smallest numerically balanced number strictly greater than 3000.
</pre>

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

<ul>
	<li><code>0 &lt;= n &lt;= 10<sup>6</sup></code></li>
</ul>
Hash Table; Math; Backtracking; Counting; Enumeration
class Solution:
    def nextBeautifulNumber(self, n: int) -> int:
        for x in count(n + 1):
            y = x
            cnt = [0] * 10
            while y:
                y, v = divmod(y, 10)
                cnt[v] += 1
            if all(v == 0 or i == v for i, v in enumerate(cnt)):
                return x
