<p>The chess knight has a <strong>unique movement</strong>,&nbsp;it may move two squares vertically and one square horizontally, or two squares horizontally and one square vertically (with both forming the shape of an <strong>L</strong>). The possible movements of chess knight are shown in this diagram:</p>

<p>A chess knight can move as indicated in the chess diagram below:</p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/18/chess.jpg" style="width: 402px; height: 402px;" />
<p>We have a chess knight and a phone pad as shown below, the knight <strong>can only stand on a numeric cell</strong>&nbsp;(i.e. blue cell).</p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/18/phone.jpg" style="width: 242px; height: 322px;" />
<p>Given an integer <code>n</code>, return how many distinct phone numbers of length <code>n</code> we can dial.</p>

<p>You are allowed to place the knight <strong>on any numeric cell</strong> initially and then you should perform <code>n - 1</code> jumps to dial a number of length <code>n</code>. All jumps should be <strong>valid</strong> knight jumps.</p>

<p>As the answer may be very large, <strong>return the answer modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 10
<strong>Explanation:</strong> We need to dial a number of length 1, so placing the knight over any numeric cell of the 10 cells is sufficient.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> 20
<strong>Explanation:</strong> All the valid number we can dial are [04, 06, 16, 18, 27, 29, 34, 38, 40, 43, 49, 60, 61, 67, 72, 76, 81, 83, 92, 94]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 3131
<strong>Output:</strong> 136006598
<strong>Explanation:</strong> Please take care of the mod.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 5000</code></li>
</ul>

class Solution {
public: 
  int knightDialer(int n) {
    vector < vector < vector < int >>> dialerComb(n, vector < vector < int >> (4, vector < int > (3, 0)));
    int mod = 1000000007;
    for (int r = 0; r < 4; r++) {
      fill(dialerComb[0][r].begin(), dialerComb[0][r].end(), 1);
    }
    for(int movNo = 0; movNo < n; movNo++){
            dialerComb[movNo][3][0] = dialerComb[movNo][3][2] = -1;
        }
        int dirs[8][2] = {{2, 1}, {2, -1}, {-2, 1}, {-2, -1}, {1, 2}, {1, -2}, {-1, 2}, {-1, -2}};
        for(int movNo = 1; movNo < n; movNo++){
            for(int r = 0; r < 4; r++){
                for(int c = 0; c < 3; c++){
                    if (dialerComb[movNo][r][c] == -1)continue;
                    for(auto dir : dirs){
                        int prevRow = r + dir[0], prevCol = c + dir[1];
                        if(isSafe(dialerComb, prevRow, prevCol, movNo-1)){
                             dialerComb[movNo][r][c] = (dialerComb[movNo][r][c] % mod +  dialerComb[movNo-1][prevRow][prevCol] % mod)%mod;
                        }
                    }
                }
            }
        }
        int unqPhnNum = 0;
        for(int r = 0; r < 4; r++){
            for(int c = 0; c < 3; c++){
                 if(dialerComb[n-1][r][c] == -1)continue;
                 unqPhnNum =  (unqPhnNum % mod + dialerComb[n-1][r][c] % mod)%mod;
            }
        }
        return unqPhnNum;
    }
    bool isSafe(vector < vector < vector < int >>> & dialerComb, int prevRow, int prevCol, int movNo){
        return movNo  > -1 && prevRow < 4 && prevCol < 3  && prevRow > -1 && prevCol > -1 && dialerComb[movNo][prevRow][prevCol] != -1;
    }
};
91202