<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:
    int n;
    int k;
    vector<string> ops;
    vector<int> numsSameConsecDiff(int n, int k) {
        this->n = n;
        this->k = k;
        string cur = "";
        ops = {"0", "1", "2", "3", "4", "5", "6", "7", "8", "9"};
        vector<int> ans;
        for(int i = 1; i < ops.size(); i++) {
            // string init = ops[i];
            cur+=ops[i];
            backtrack(cur, ans);
            cur.pop_back();
        }
        // backtrack(cur, ans);
        // prune(ans);
        return ans;
    }
    
    void backtrack (string cur, vector<int>& ans) {
        //base case
        if(cur.length() == n) {
            ans.push_back(stoi(cur));
            return;
        }
        
        int digit = cur.back() - '0';
        // vector<int> options = {digit - k, digit + k}
        if(digit - k == digit + k) {
            cur+=ops[digit-k];
            backtrack(cur, ans);
            cur.pop_back();
        } else {
            if(digit - k >= 0) {
                cur+=ops[digit-k];
                backtrack(cur, ans);
                cur.pop_back();
            }
            if(digit + k <= 9) {
                cur+=ops[digit+k];
                backtrack(cur, ans);
                cur.pop_back();
            }    
        }
        
    }
};
12500