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[10] = {};
        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

All Problems

All Solutions