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 {
public:
    vector<int> numsSameConsecDiff(int n, int k) {
        vector<int> ans;
        int boundary = pow(10, n - 1);
        auto dfs = [&](this auto&& dfs, int x) {
            if (x >= boundary) {
                ans.push_back(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);
            }
        };
        for (int i = 1; i < 10; ++i) {
            dfs(i);
        }
        return ans;
    }
};
