<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:
    // void dfs(int last, int length, int n, int k, vector<int> &ans, string s){
    //     if(s.length()==n){
    //         ans.push_back(stoi(s));
    //         return;
    //     }
    //     if(last + k <= 9){
    //         s += to_string(last+k);
    //         dfs(last + k,length+1,n,k,ans,s);
    //         s.pop_back();
    //     }

    //     if(k != 0){
    //         if(last - k >= 0){
    //             s += to_string(last-k);
    //             dfs(last - k,length+1,n,k,ans,s);
    //             s.pop_back();
    //         }  
    //     }  
    // }
    vector<int> numsSameConsecDiff(int n, int k) {

        //DFS
        // vector<int>ans;
        // for(int i=1;i<=9;i++){
        //     string s = to_string(i);
        //     dfs(i,1,n,k,ans,s);
        // }
        // return ans;

        //BFS ->MORE EASIER APPROACH
        vector<int>curr = {1,2,3,4,5,6,7,8,9};
        for(int i = 2; i<= n;i++){
            vector<int>curr2;
            for(int x : curr){
                int dig = x % 10;
                if(dig + k <= 9){
                    curr2.push_back(x*10+dig+k);
                }
                if(k > 0 && dig - k >= 0){
                    curr2.push_back(x*10+dig-k);
                }
            }
            curr = curr2;
        }
        return curr;
    }
};
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