Paths in Matrix Whose Sum Is Divisible by K
<p>You are given a <strong>0-indexed</strong> <code>m x n</code> integer matrix <code>grid</code> and an integer <code>k</code>. You are currently at position <code>(0, 0)</code> and you want to reach position <code>(m - 1, n - 1)</code> moving only <strong>down</strong> or <strong>right</strong>.</p>

<p>Return<em> the number of paths where the sum of the elements on the path is divisible by </em><code>k</code>. Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2400-2499/2435.Paths%20in%20Matrix%20Whose%20Sum%20Is%20Divisible%20by%20K/images/image-20220813183124-1.png" style="width: 437px; height: 200px;" />
<pre>
<strong>Input:</strong> grid = [[5,2,4],[3,0,5],[0,7,2]], k = 3
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are two paths where the sum of the elements on the path is divisible by k.
The first path highlighted in red has a sum of 5 + 2 + 4 + 5 + 2 = 18 which is divisible by 3.
The second path highlighted in blue has a sum of 5 + 3 + 0 + 5 + 2 = 15 which is divisible by 3.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2400-2499/2435.Paths%20in%20Matrix%20Whose%20Sum%20Is%20Divisible%20by%20K/images/image-20220817112930-3.png" style="height: 85px; width: 132px;" />
<pre>
<strong>Input:</strong> grid = [[0,0]], k = 5
<strong>Output:</strong> 1
<strong>Explanation:</strong> The path highlighted in red has a sum of 0 + 0 = 0 which is divisible by 5.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2400-2499/2435.Paths%20in%20Matrix%20Whose%20Sum%20Is%20Divisible%20by%20K/images/image-20220812224605-3.png" style="width: 257px; height: 200px;" />
<pre>
<strong>Input:</strong> grid = [[7,3,4,9],[2,3,6,2],[2,3,7,0]], k = 1
<strong>Output:</strong> 10
<strong>Explanation:</strong> Every integer is divisible by 1 so the sum of the elements on every possible path is divisible by k.
</pre>

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

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= m * n &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= grid[i][j] &lt;= 100</code></li>
	<li><code>1 &lt;= k &lt;= 50</code></li>
</ul>
Array; Dynamic Programming; Matrix
function numberOfPaths(grid: number[][], k: number): number {
    const MOD = 10 ** 9 + 7;
    const m = grid.length,
        n = grid[0].length;
    let ans = Array.from({ length: m + 1 }, () =>
        Array.from({ length: n + 1 }, () => new Array(k).fill(0)),
    );
    ans[0][1][0] = 1;
    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            for (let v = 0; v < k; v++) {
                let key = (grid[i][j] + v) % k;
                ans[i + 1][j + 1][key] =
                    (ans[i][j + 1][v] + ans[i + 1][j][v] + ans[i + 1][j + 1][key]) % MOD;
            }
        }
    }
    return ans[m][n][0];
}
