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LeetCode 1215: Stepping Numbers Solution

Master LeetCode problem 1215 (Stepping Numbers), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1215. Stepping Numbers

Medium
MathBacktrackingBreadth-First Search

Problem Explanation

Explanation:

Given a range [low, high], we need to find all stepping numbers in that range. A stepping number is a number where each digit differs by 1 from the adjacent digits.

To solve this problem, we will perform a Breadth First Search (BFS) starting from the single-digit numbers (1 to 9) and expanding to the next possible stepping numbers in a level-wise manner. We will keep track of all valid stepping numbers within the given range.

Algorithm:

  1. Initialize a queue with all single-digit numbers from 0 to 9.
  2. While the queue is not empty, pop a number from the queue.
  3. If the number is within the range [low, high], add it to the result.
  4. For each valid stepping number, i.e., numbers where the last digit is at most 1 away from the current number:
    • If the new number is less than or equal to high, add it to the queue to explore further.
  5. Repeat steps 2-4 until the queue is empty.
  6. Return the list of stepping numbers within the range [low, high].

Time Complexity:

  • The time complexity of this algorithm is O(n), where n is the total number of stepping numbers within the given range.

Space Complexity:

  • The space complexity of this algorithm is O(k), where k is the total number of stepping numbers within the given range.

:

Solution Code

class Solution {
    public List<Integer> countSteppingNumbers(int low, int high) {
        List<Integer> result = new ArrayList<>();
        Queue<Integer> queue = new LinkedList<>();
        
        if (low == 0) {
            result.add(0);
        }
        
        for (int i = 1; i <= 9; i++) {
            queue.offer(i);
        }
        
        while (!queue.isEmpty()) {
            int curr = queue.poll();
            if (curr >= low && curr <= high) {
                result.add(curr);
            }
            int lastDigit = curr % 10;
            
            if (lastDigit > 0) {
                int nextNum = curr * 10 + (lastDigit - 1);
                if (nextNum <= high) {
                    queue.offer(nextNum);
                }
            }
            
            if (lastDigit < 9) {
                int nextNum = curr * 10 + (lastDigit + 1);
                if (nextNum <= high) {
                    queue.offer(nextNum);
                }
            }
        }
        
        return result;
    }
}

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