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2435. Paths in Matrix Whose Sum Is Divisible by K.py

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+# 2435. Paths in Matrix Whose Sum Is Divisible by K
+
+# You are given a 0-indexed m x n integer matrix grid and an integer k. You are currently at position (0, 0) and you want to reach position (m - 1, n - 1) moving only down or right.
+
+# Return the number of paths where the sum of the elements on the path is divisible by k. Since the answer may be very large, return it modulo 109 + 7.
+
+ 
+
+# Example 1:
+
+
+# Input: grid = [[5,2,4],[3,0,5],[0,7,2]], k = 3
+# Output: 2
+# Explanation: 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.
+# Example 2:
+
+
+# Input: grid = [[0,0]], k = 5
+# Output: 1
+# Explanation: The path highlighted in red has a sum of 0 + 0 = 0 which is divisible by 5.
+# Example 3:
+
+
+# Input: grid = [[7,3,4,9],[2,3,6,2],[2,3,7,0]], k = 1
+# Output: 10
+# Explanation: Every integer is divisible by 1 so the sum of the elements on every possible path is divisible by k.
+ 
+
+# Constraints:
+
+# m == grid.length
+# n == grid[i].length
+# 1 <= m, n <= 5 * 104
+# 1 <= m * n <= 5 * 104
+# 0 <= grid[i][j] <= 100
+# 1 <= k <= 50
+
+
+class Solution:
+    def numberOfPaths(self, grid, k):
+        MOD = 10**9 + 7
+        m, n = len(grid), len(grid[0])
+        
+        # dp[i][j][mod] = ways to reach (i,j) with sum % k = mod
+        dp = [[[0] * k for _ in range(n)] for _ in range(m)]
+        dp[0][0][grid[0][0] % k] = 1
+        
+        for i in range(m):
+            for j in range(n):
+                for mod in range(k):
+                    if dp[i][j][mod] > 0:
+                        if i + 1 < m:
+                            new_mod = (mod + grid[i+1][j]) % k
+                            dp[i+1][j][new_mod] = (dp[i+1][j][new_mod] + dp[i][j][mod]) % MOD
+                        if j + 1 < n:
+                            new_mod = (mod + grid[i][j+1]) % k
+                            dp[i][j+1][new_mod] = (dp[i][j+1][new_mod] + dp[i][j][mod]) % MOD
+        
+        return dp[m-1][n-1][0]