How do you solve Count Stepping Numbers in Range on LeetCode?

Count Stepping Numbers in Range (LeetCode #2801) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Stepping numbers are defined by the absolute difference of adjacent digits being 1.

What is the brute force solution for Count Stepping Numbers in Range?

The brute force approach for Count Stepping Numbers in Range is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all numbers from 1 to high and checking each number to see if it is a stepping number. This is straightforward but inefficient for large ranges. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Stepping Numbers in Range?

Count Stepping Numbers in Range uses the following concepts: String, Dynamic Programming. The recommended patterns to study are: BFS/DFS for tree/graph traversal, Dynamic programming for state-based problems.

What are the interview tips for Count Stepping Numbers in Range?

Always clarify the constraints and edge cases with the interviewer. Explain your thought process clearly while coding. Consider both brute-force and optimized solutions to demonstrate your understanding.

What are common mistakes when solving Count Stepping Numbers in Range?

Not considering leading zeros when generating numbers. Failing to handle large numbers due to constraints.

#2801

Count Stepping Numbers in Range

Hard

StringDynamic ProgrammingBFS/DFS for tree/graph traversalDynamic programming for state-based problems

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Instead of checking every number, we can use a breadth-first search (BFS) or depth-first search (DFS) to generate only the stepping numbers. This drastically reduces the number of checks we need to perform.

⚙️

Algorithm

3 steps

1Step 1: Use BFS or DFS to generate stepping numbers starting from each digit (1-9).
2Step 2: For each generated number, check if it lies within the range [low, high].
3Step 3: Count valid stepping numbers and return the count.

solution.py20 lines

1# Full working Python code
2class Solution:
3    def countSteppingNumbers(self, low: str, high: str) -> int:
4        low, high = int(low), int(high)
5        count = 0
6        for start in range(1, 10):
7            self.bfs(start, low, high, count)
8        return count % (10**9 + 7)
9
10    def bfs(self, start, low, high, count):
11        queue = [start]
12        while queue:
13            num = queue.pop(0)
14            if low <= num <= high:
15                count += 1
16            last_digit = num % 10
17            if last_digit > 0:
18                queue.append(num * 10 + (last_digit - 1))
19            if last_digit < 9:
20                queue.append(num * 10 + (last_digit + 1))

ℹ

Complexity note: The time complexity is O(n) because we only generate valid stepping numbers rather than checking every number in the range, which is much more efficient.

1Stepping numbers are defined by the absolute difference of adjacent digits being 1.
2Using BFS/DFS allows us to generate only valid stepping numbers, improving efficiency.

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