<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://assets.leetcode.com/uploads/2022/08/13/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://assets.leetcode.com/uploads/2022/08/17/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://assets.leetcode.com/uploads/2022/08/12/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>

from collections import defaultdict
from typing import Dict, Tuple

Cell = Tuple[int, int]
Counts = Dict[int, int]

class PathSolver:
    def __init__(self, grid: List[List[int]]):
        self.grid = grid
        self.num_rows = len(grid)
        self.num_cols = len(grid[0])

    def init_dp(self, k: int) -> Dict[Cell, Counts]:
        dp = {(r, c): defaultdict(int) for r in range(self.num_rows) for c in range(self.num_cols)}
        dp[(0, 0)][self.grid[0][0] % k] += 1
        return dp

    def get_path_indices(self, path_length: int) -> List[Cell]:
        r = min(path_length, self.num_rows) - 1
        c = path_length - 1 - r
        idx = []
        cell = (r, c)
        while self.in_bounds(cell):
            idx.append(cell)
            cell = (cell[0] - 1, cell[1] + 1)
        return idx

    def in_bounds(self, cell: Tuple[int, int]) -> bool:
        return (0 <= cell[0] < self.num_rows) and (0 <= cell[1] < self.num_cols)

    def get_previous_neighbors(self, cell: Cell) -> List[Cell]:
        neighbors = []
        for r_offset, c_offset in [(-1, 0), (0, -1)]:
            prev = (cell[0] + r_offset, cell[1] + c_offset)
            if self.in_bounds(prev):
                neighbors.append(prev)
        return neighbors

    def solve(self, k: int) -> int:
        dp = self.init_dp(k)
        for path_length in range(2, self.num_rows + self.num_cols):
            for r, c in self.get_path_indices(path_length):
                for prev_neighbor in self.get_previous_neighbors((r, c)):
                    for prev_mod, prev_count in dp[prev_neighbor].items():
                        dp[(r, c)][(prev_mod + self.grid[r][c]) % k] += prev_count
        return dp[(self.num_rows - 1, self.num_cols - 1)].get(0, 0)


class Solution:
    def numberOfPaths(self, grid: List[List[int]], k: int) -> int:
        return PathSolver(grid).solve(k) % (10 ** 9 + 7)

179327