How do you solve Next Greater Numerically Balanced Number on LeetCode?

Next Greater Numerically Balanced Number (LeetCode #2048) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(n) space complexity. Key insight: Numerically balanced numbers are limited to digits 1-9, as 0 cannot appear.

What is the brute force solution for Next Greater Numerically Balanced Number?

The brute force approach for Next Greater Numerically Balanced Number is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each number greater than n to see if it is numerically balanced. This is straightforward but inefficient, as it may require checking many numbers before finding a valid one. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Next Greater Numerically Balanced Number?

Next Greater Numerically Balanced Number uses the following concepts: Hash Table, Math, Backtracking, Counting, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Next Greater Numerically Balanced Number?

Always clarify the constraints and edge cases before diving into the solution. Consider generating possible solutions rather than checking each number sequentially. Practice recognizing patterns in problems to optimize your approach.

What are common mistakes when solving Next Greater Numerically Balanced Number?

Not considering that 0 cannot appear in a numerically balanced number. Failing to check all digits correctly for their counts.

#2048

Next Greater Numerically Balanced Number

Medium

Hash TableMathBacktrackingCountingEnumerationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(1)
Space	O(1)	O(n)

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Intuition

Time O(1)Space O(n)

The optimal solution involves generating numerically balanced numbers directly instead of checking each number. We can build these numbers based on the properties of numerically balanced numbers, ensuring we find the smallest one greater than n efficiently.

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Algorithm

4 steps

1Step 1: Generate numerically balanced numbers by iterating through possible digit counts (1 to 9).
2Step 2: For each digit d, create the number by repeating d exactly d times.
3Step 3: Store these generated numbers and sort them.
4Step 4: Find the smallest number in the sorted list that is greater than n.

solution.py17 lines

1# Full working Python code
2
3def generate_balanced_numbers():
4    balanced_numbers = []
5    for d in range(1, 10):
6        num = int(str(d) * d)
7        balanced_numbers.append(num)
8    return sorted(balanced_numbers)
9
10def next_greater_balanced(n):
11    balanced_numbers = generate_balanced_numbers()
12    for num in balanced_numbers:
13        if num > n:
14            return num
15
16# Example usage
17print(next_greater_balanced(1))  # Output: 22

ℹ

Complexity note: The time complexity is O(1) for finding the next greater number because we are generating a fixed set of numerically balanced numbers (only 9 possible digits), and the search is constant time. The space complexity is O(n) due to storing the generated numbers.

1Numerically balanced numbers are limited to digits 1-9, as 0 cannot appear.
2The maximum numerically balanced number under the constraints is 999999.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.