LeetCode 2801: Count Stepping Numbers in Range Solution

Master LeetCode problem 2801 (Count Stepping Numbers in Range), a hard 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.

2801. Count Stepping Numbers in Range

Hard

String Dynamic Programming

Problem Explanation

Explanation:

We can solve this problem using a breadth-first search (BFS) approach.
Starting from each digit from 1 to 9, we perform a BFS traversal by considering all possible next stepping numbers.
For each stepping number, we check if it falls within the range [low, high].
We continue the BFS traversal until we exhaust all stepping numbers within the given range.
Finally, we return the count of stepping numbers modulo 10^9 + 7. Solution:

Solution Code

class Solution {
    public int countSteppingNumbers(String low, String high) {
        long mod = 1000000007;
        long result = 0;
        long l = Long.parseLong(low);
        long h = Long.parseLong(high);
        Queue<Long> queue = new LinkedList<>();
        for (long i = 1; i <= 9; i++) {
            queue.offer(i);
        }
        if (l == 0) {
            result++;
        }
        
        while (!queue.isEmpty()) {
            long num = queue.poll();
            if (num >= l && num <= h) {
                result++;
            }
            if (num > h) {
                continue;
            }
            long lastDigit = num % 10;
            if (lastDigit > 0) {
                queue.offer(num * 10 + lastDigit - 1);
            }
            if (lastDigit < 9) {
                queue.offer(num * 10 + lastDigit + 1);
            }
        }
        
        return (int) (result % mod);
    }
}

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