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
class Solution {
    private int m;
    private int n;
    private int k;
    private static final int MOD = (int) 1e9 + 7;
    private int[][] grid;
    private int[][][] f;

    public int numberOfPaths(int[][] grid, int k) {
        this.grid = grid;
        this.k = k;
        m = grid.length;
        n = grid[0].length;
        f = new int[m][n][k];
        for (var a : f) {
            for (var b : a) {
                Arrays.fill(b, -1);
            }
        }
        return dfs(0, 0, 0);
    }

    private int dfs(int i, int j, int s) {
        if (i < 0 || i >= m || j < 0 || j >= n) {
            return 0;
        }
        s = (s + grid[i][j]) % k;
        if (f[i][j][s] != -1) {
            return f[i][j][s];
        }
        if (i == m - 1 && j == n - 1) {
            return s == 0 ? 1 : 0;
        }
        int ans = dfs(i + 1, j, s) + dfs(i, j + 1, s);
        ans %= MOD;
        f[i][j][s] = ans;
        return ans;
    }
}