Formatted question description: https://leetcode.ca/all/2048.html

# 2048. Next Greater Numerically Balanced Number (Medium)

An integer x is numerically balanced if for every digit d in the number x, there are exactly d occurrences of that digit in x.

Given an integer n, return the smallest numerically balanced number strictly greater than n.

Example 1:

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


Example 2:

Input: n = 1000
Output: 1333
Explanation:
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.


Example 3:

Input: n = 3000
Output: 3133
Explanation:
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.


Constraints:

• 0 <= n <= 106

## Solution 1. Brute Force

// OJ: https://leetcode.com/problems/next-greater-numerically-balanced-number/
// Time: O(C * logC) where C is the maximum possible input number.
// Space: O(1)
class Solution {
bool balance(int n) {
int cnt = {};
while (n) {
if (n % 10 == 0) return false; // no 0 allowed
cnt[n % 10]++;
n /= 10;
}
for (int i = 1; i < 10; ++i) {
if (cnt[i] && cnt[i] != i) return false;
}
return true;
}
public:
int nextBeautifulNumber(int n) {
while (true) {
++n;
if (balance(n)) return n;
}
}
};


## Discuss

https://leetcode.com/problems/next-greater-numerically-balanced-number/discuss/1537491/C%2B%2B-Brute-Force