LeetCode #2048 — MEDIUM

Next Greater Numerically Balanced Number

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Math · Backtracking

Example 1

Example 2

Example 3

Core Insight

What unlocks the optimal approach

How far away can the next greater numerically balanced number be from n?
With the given constraints, what is the largest numerically balanced number?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2048: Next Greater Numerically Balanced Number
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;
            }
        }
    }
}

// Accepted solution for LeetCode #2048: Next Greater Numerically Balanced Number
func nextBeautifulNumber(n int) int {
	for x := n + 1; ; x++ {
		cnt := [10]int{}
		for y := x; y > 0; y /= 10 {
			cnt[y%10]++
		}
		ok := true
		for y := x; y > 0; y /= 10 {
			if y%10 != cnt[y%10] {
				ok = false
				break
			}
		}
		if ok {
			return x
		}
	}
}

# Accepted solution for LeetCode #2048: Next Greater Numerically Balanced Number
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

// Accepted solution for LeetCode #2048: Next Greater Numerically Balanced Number
impl Solution {
    pub fn next_beautiful_number(n: i32) -> i32 {
        let mut x = n + 1;
        loop {
            let mut cnt = [0; 10];
            let mut y = x;
            while y > 0 {
                cnt[(y % 10) as usize] += 1;
                y /= 10;
            }
            let mut ok = true;
            let mut y2 = x;
            while y2 > 0 {
                let d = (y2 % 10) as usize;
                if d != cnt[d] {
                    ok = false;
                    break;
                }
                y2 /= 10;
            }
            if ok {
                return x;
            }
            x += 1;
        }
    }
}

// Accepted solution for LeetCode #2048: Next Greater Numerically Balanced Number
function nextBeautifulNumber(n: number): number {
    for (let x = n + 1; ; ++x) {
        const cnt: number[] = Array(10).fill(0);
        for (let y = x; y > 0; y = (y / 10) | 0) {
            cnt[y % 10]++;
        }
        let ok = true;
        for (let i = 0; i < 10; ++i) {
            if (cnt[i] && cnt[i] !== i) {
                ok = false;
                break;
            }
        }
        if (ok) {
            return x;
        }
    }
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n!)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.