<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) {
        unordered_set<string> vis;

        vis.insert("");

        queue<string> q;
        q.push("");
        vector<int> op;
        while(!q.empty()) {
            auto curr = q.front(); q.pop();

            if (curr.length() == n) {
                op.push_back(stoi(curr));
            } else if (curr.length() == 0) {
                for (int i = 1; i <= 9; i++) {
                    q.push(curr+(char)(i+'0'));
                    vis.insert(curr+(char)(i+'0'));
                }
            }
            else {
                int last = curr[curr.length()-1]-'0';

                if (last - k >= 0) {
                    curr.push_back(last-k+'0');  

                    if (!contains(vis, curr)) {
                        vis.insert(curr);
                        q.push(curr);
                    }

                    curr.pop_back();

                }

                
                

                if (last + k <= 9) {
                    curr.push_back(last+k+'0');

                    if (!contains(vis, curr)) {
                        vis.insert(curr);
                        q.push(curr);
                    }

                    curr.pop_back();
                }

                
            }
        }

        return op;
    }

    bool contains(unordered_set<string>& st, string& ele) {
        return st.find(ele) != st.end();
    }
};
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