<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>

class Solution:
    def nextBeautifulNumber(self, n: int) -> int:
        def backtracking(i, numLen, curNum, counter):
            if i == numLen:
                isBalanceNumber = True
                for d, freq in counter.items():
                    if freq != 0 and d != freq:
                        isBalanceNumber = False
                if isBalanceNumber:
                    yield curNum
                return

            for d in range(1, numLen+1):
                if counter[d] >= d: continue   # Prune if number of occurrences of `d` is greater than `d`
                if counter[d] + (numLen - i) < d: continue  # Prune if not enough number of occurrences of `d`
                counter[d] += 1
                yield from backtracking(i + 1, numLen, curNum * 10 + d, counter)
                counter[d] -= 1

        for numLen in range(len(str(n)), len(str(n)) + 2):
            for num in backtracking(0, numLen, 0, Counter()):
                if num > n:
                    return num
16300