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 {
    public int nextBeautifulNumber(int n) {
        for (int x = n + 1;; ++x) {
            int[] cnt = new int[10];
            for (int y = x; y > 0; y /= 10) {
                ++cnt[y % 10];
            }
            boolean ok = true;
            for (int y = x; y > 0; y /= 10) {
                if (y % 10 != cnt[y % 10]) {
                    ok = false;
                    break;
                }
            }
            if (ok) {
                return x;
            }
        }
    }
}