LeetCode 967: Numbers With Same Consecutive Differences Solution

Master LeetCode problem 967 (Numbers With Same Consecutive Differences), 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.

967. Numbers With Same Consecutive Differences

Medium

Backtracking Breadth-First Search

Problem Explanation

Explanation

To solve this problem, we can use a breadth-first search (BFS) approach. We start with single digit numbers (1 to 9) and then iteratively append a digit to the right of each number based on the given difference k. We continue this process until we reach the desired length n. We maintain a queue to store the current numbers being processed.

Algorithm

Initialize a queue with single digit numbers from 1 to 9.
While the length of the numbers in the queue is less than n, do the following:
- Pop a number from the queue.
- Get the last digit of the number.
- Generate new numbers by adding/subtracting k to the last digit (if the result is between 0 and 9).
- Append the new numbers to the queue.
Return the numbers in the queue.

Time Complexity

The time complexity of this algorithm is O(2^N) where N is the given length of the numbers.

Space Complexity

The space complexity of this algorithm is O(2^N) as well, as we store the generated numbers in a queue.

Solution Code

import java.util.*;

class Solution {
    public int[] numsSameConsecDiff(int n, int k) {
        Queue<Integer> queue = new LinkedList<>();
        for (int i = 1; i <= 9; i++) {
            queue.offer(i);
        }

        while (queue.peek().toString().length() < n) {
            int num = queue.poll();
            int lastDigit = num % 10;
            if (lastDigit + k <= 9) {
                queue.offer(num * 10 + lastDigit + k);
            }
            if (lastDigit - k >= 0 && k != 0) {
                queue.offer(num * 10 + lastDigit - k);
            }
        }

        return queue.stream().mapToInt(Integer::intValue).toArray();
    }
}

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