<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>

class Solution {
public:
    vector<int> numsSameConsecDiff(int n, int k) {
        vector<int>ans;
        unordered_set<int>s;
        queue<string>q;
        for (int i = 1; i < 10; i++) {
            q.push(to_string(i));
        }
        while (!q.empty()) {
            string curr = q.front();
            q.pop();
            if (curr.size() == n) {
                s.insert(stoi(curr));
                continue;
            }
            int d = curr[curr.size()-1] - '0';
            if ((d + k >= 0) && (d + k <= 9)) q.push(curr + to_string(d+k));
            if ((d - k >= 0) && (d - k <= 9)) q.push(curr + to_string(d-k));
        }
        for (auto x : s) ans.push_back(x);
        return ans;
    }
};
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