<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>

class Solution:
    def numberOfPaths(self, grid: List[List[int]], k: int) -> int:
        m, n = len(grid), len(grid[0])
        
        # We are trying to implement DP where we traverse along the columns
        # Hence, the size of DP list is m * k
        # Hence if m, n, operate on the transpose
        if m > n:
            return self.numberOfPaths([[grid[r][c] for r in range(m)] for c in range(n)], k)

        MOD = 10 ** 9 + 7

        # A = [a_i] where ai = number of pathSum that has value i after modulo with k
        # If we add r to the array, it is equivalent to bring the last r elements to the front
        # For eg k = 3, r = 1, A = [5, 1, 7]. Since we add 1, a2 -> a0, a1 -> a2 and a0 -> a1
        def rotate(A, r):
            if r == 0:
                return A
            return A[-r:] + A[:-r]

        # Add to list A, B element wise and then rotate
        def addRotate(A, B, r):
            x = [(a + b) % MOD for a, b in zip(A, B)]
            return rotate(x, r)

        # Initially DP[i] is a list that represent number of pathSums with modulo k starting 
        # from grid[-1][row] to grid[-1][-1] 
        DP = [[0] * k for _ in range(m)]

        # Update for the last element
        DP[-1][grid[-1][-1] % k] = 1

        # Handle the first column separately
        for row in range(m - 2, -1, -1):
            DP[row] = rotate(DP[row + 1], grid[row][-1] % k)

        # Iterate over col from right to left
        for col in range(n - 2, -1, -1):
            # Update the last element separately as DP[row][col] only depends on DP[row][col + 1]
            DP[-1] = rotate(DP[-1], grid[-1][col] % k)

            for row in range(m - 2, -1, -1):
                # path from (r, c) passes through either (r + 1, c) or (r, c + 1)
                # Since (r, .) depends on on (>=r, .), we can optimze space and do it in one dimension
                DP[row] = addRotate(DP[row], DP[row + 1], grid[row][col] % k)
        
        return DP[0][0]
951