LeetCode Entry

2048. Next Greater Numerically Balanced Number

24.10.2025 medium 2025 kotlin rust

First bigger number freq[digit]=digit

2048. Next Greater Numerically Balanced Number medium blog post substack youtube

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Problem TLDR

First bigger number freq[digit]=digit #medium

Intuition

Brute-force is accepted. For the problem size of 10^6 the next value is 1224444 which is just 200k loop.

Approach

• or we can try to generate all permutations; prune by length of the initial number

Complexity

• Time complexity: \(O(n)\)

• Space complexity: \(O(1)\)

Code

// 425ms
    fun nextBeautifulNumber(n: Int) = (n+1..n*21+1)
        .first { "$it".groupBy { it }.all { it.key-'0' == it.value.size }}

// 22ms
    pub fn next_beautiful_number(n: i32) -> i32 {
        (n+1..n*22+2).find(|&x| { let (mut y,mut f) = (x, [0;10]);
            while y > 0 { f[(y%10)as usize] += 1; y /= 10 }
            (0..=9).all(|x| f[x] == x || f[x] < 1)
        }).unwrap()
    }