Numbers With Same Consecutive Differences
<p>Given two integers n and k, return <em>an array of all the integers of length </em><code>n</code><em> where the difference between every two consecutive digits is </em><code>k</code>. You may return the answer in <strong>any order</strong>.</p>

<p>Note that the integers should not have leading zeros. Integers as <code>02</code> and <code>043</code> are not allowed.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 3, k = 7
<strong>Output:</strong> [181,292,707,818,929]
<strong>Explanation:</strong> Note that 070 is not a valid number, because it has leading zeroes.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 2, k = 1
<strong>Output:</strong> [10,12,21,23,32,34,43,45,54,56,65,67,76,78,87,89,98]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n &lt;= 9</code></li>
	<li><code>0 &lt;= k &lt;= 9</code></li>
</ul>
Breadth-First Search; Backtracking
class Solution {
    private List<Integer> ans = new ArrayList<>();
    private int boundary;
    private int k;

    public int[] numsSameConsecDiff(int n, int k) {
        this.k = k;
        boundary = (int) Math.pow(10, n - 1);
        for (int i = 1; i < 10; ++i) {
            dfs(i);
        }
        return ans.stream().mapToInt(i -> i).toArray();
    }

    private void dfs(int x) {
        if (x >= boundary) {
            ans.add(x);
            return;
        }
        int last = x % 10;
        if (last + k < 10) {
            dfs(x * 10 + last + k);
        }
        if (k != 0 && last - k >= 0) {
            dfs(x * 10 + last - k);
        }
    }
}
