2435. Paths in Matrix Whose Sum Is Divisible by K

Problem Description

In this LeetCode problem, we have a two-dimensional grid representing a matrix with m rows and n columns, and we are tasked with finding paths from the top left corner (0, 0) to the bottom right corner (m - 1, n - 1). The only permitted movements are either right or down. Each cell in the grid contains an integer, and we want to consider only those paths for which the sum of the integers along the path is divisible by a given integer k.

The problem statement asks us to return the total number of such paths modulo 10^9 + 7. This large number is used to prevent overflow issues due to very large result values, which is a common practice in computational problems.

Intuition

To arrive at the solution, we have to think in terms of dynamic programming, which is a method for solving problems by breaking them down into simpler subproblems. The main idea behind the approach is to create a 3-dimensional array dp where each element dp[i][j][s] represents the number of ways to reach cell (i, j) such that the sum of all elements in the path modulo k is s.

Here's how we can think through it:

1. Initialize a 3-dimensional array dp of size m x n x k with zeroes, which will store the count of paths that lead to a certain remainder when the sum of path elements is divided by k. The third dimension s represents all possible remainders [0, k-1].

2. Set dp[0][0][grid[0][0] % k] to 1 as a base case since there's one way to be at the starting cell with the sum equal to the element of that cell modulo k.

3. Start iterating over the grid, cell by cell. At each cell, we want to update the dp array for all possible sums modulo k. We consider two possibilities to arrive at a cell (i, j): from the cell above (i - 1, j) and from the cell to the left (i, j - 1). For each of these cells, we add to the path count for the current remainder s.

4. We calculate the new remainder t after including the current cell's value using the formula t = ((s - grid[i][j] % k) + k) % k. This gives us the remainder from the previous cell that would lead to a current remainder s after adding grid[i][j].

5. If the cell above (i - 1, j) is valid, we add the number of ways to reach it with the remainder t to dp[i][j][s]. If the cell to the left (i, j - 1) is valid, we do the same.

6. Since we only care about the counts modulo 10^9 + 7, we take the modulo at each update step to keep numbers in the range.

7. Finally, we're interested in the number of paths that have a sum divisible by k when we've reached the bottom-right cell. This corresponds to dp[m - 1][n - 1][0], the number of paths with a remainder of 0, which we return as the answer.

By following these steps, we can fill up our dp table and compute the required value efficiently, avoiding the need to explicitly enumerate all possible paths, which would be impractical for large grids.

Learn more about Dynamic Programming patterns.

Solution Approach

The implementation utilizes dynamic programming to efficiently compute the number of paths that lead to the bottom-right corner of the grid with sums divisible by k:

1. Initialization: We initialize a 3D list, dp, of size m * n * k, where m is the number of rows and n is the number of columns in the grid. This array will store the count of paths for each possible sum modulo k (s ranges from 0 to k-1) for each cell. Initially, all elements are set to 0.

2. Base Case: The path count of the starting position (0, 0) is set such that dp[0][0][grid[0][0] % k] is 1, because there is only one way to be at the starting cell and the sum equals the element in the starting cell modulo k.

   1dp[0][0][grid[0][0] % k] = 1

3. Main Logic: We iterate through each cell (i, j) in the grid, and for each cell, we iterate through all possible remainders s (from 0 to k-1). We update dp[i][j][s] by adding the number of ways to reach either the cell above (i - 1, j) or the cell to the left (i, j - 1) that would result in a remainder s after adding the value of the current cell.

   The transition formula used is:

   1t = ((s - grid[i][j] % k) + k) % k
   2if i:
   3    dp[i][j][s] += dp[i - 1][j][t]
   4if j:
   5    dp[i][j][s] += dp[i][j - 1][t]

   For each dp update, we take the modulo operation to ensure the result stays within the required range:

   1dp[i][j][s] %= mod

4. Final Result: The number of paths where the sum of the elements is divisible by k is the last cell's value dp[m - 1][n - 1][0], since we are interested in paths with a sum that has a remainder of 0 after modulo division by k.

   1return dp[-1][-1][0]

By following this approach, we can compute the number of valid paths without visiting all possible paths, which would be computationally expensive especially on larger grids. This solution leverages the property of modulo operation and the principle of dynamic programming, specifically memoization, to store intermediate results and avoid repetitive work. The modular arithmetic ensures that we handle large numbers efficiently and prevent arithmetic overflow, which is a common issue in problems involving counting and combinatorics on a large scale.

Example Walkthrough

Let's illustrate the dynamic programming approach with a small example. Consider a 2x3 grid with the following values:

1[
2  [1, 1, 2],
3  [2, 3, 4]
4]

And let k = 3. We want to find all the paths from the top left corner to the bottom right with sums divisible by k.

1. Initialization: We first set up a 3D array dp of size 2x3x3 (since m = 2, n = 3, and k = 3). We initialize all elements to 0.

2. Base case: For the starting cell (0, 0) with the value 1, we set dp[0][0][1 % 3] = dp[0][0][1] = 1. There is one way to be at (0, 0) with a sum that modulo k is 1.

3. Main logic:

   • Starting with cell (0, 1) with the value 1, the remainder s will be ((1 + 1) % 3) = 2. Since we can only arrive from the left, we will update dp[0][1][2] to 1.

   • Now for cell (0, 2) with the value 2, s will be ((2 + 2) % 3) = 1. Update dp[0][2][1] to 1 as we can only arrive from the left.

   • For cell (1, 0) with the value 2, s will be ((1 + 2) % 3) = 0. Update dp[1][0][0] to 1 since we can only arrive from above.

   • At cell (1, 1) with the value 3, we check from above (0, 1) and left (1, 0). From above, t = ((s - 3 % 3) + 3) % 3 = s, we add dp[0][1][t] = dp[0][1][s] to dp[1][1][s]. We will do the same for the left cell (1, 0). Hence, dp[1][1][0] will be updated to 2 (1 from above, 1 from the left).

   • Lastly, for cell (1, 2) with the value 4, the remainder s will be ((0 + 4) % 3) = 1. Update dp[1][2][1] by adding counts from above (1, 1) with value dp[1][1][t] where t = ((1 - 4 % 3) + 3) % 3 = 0, and from the left (1, 2) with value dp[1][1][1]. Thus, dp[1][2][1] will increase by 2 (all from above, nothing from the left, as dp[1][1][1] is 0).

4. Final result: We look at dp[1][2][0], but in our case, the number of paths that end with a sum divisible by k is stored in dp[1][2][1]. Since the bottom right cell ((1, 2)) has a sum remainder s = 1, not 0, there are no paths that sum up to a number divisible by k, and the method would return 0.

Using this approach, we managed to calculate the required paths without exhaustively iterating through all paths. We used a combination of iterative updates based on previous states and modular arithmetic to maintain efficiency and correctness.

Solution Implementation

Time and Space Complexity

The provided Python code defines a method to calculate the number of paths on a 2D grid where the sum of the values along the path is divisible by k. It uses dynamic programming to store the counts for intermediate paths where the sum of the values modulo k is a specific remainder.

Time Complexity:

The time complexity of the given code can be analyzed by considering the three nested loops:

1. The outermost loop runs for m iterations, where m is the number of rows in the grid.
2. The middle loop runs for n iterations for each i, where n is the number of columns in the grid.
3. The innermost loop runs for k iterations for each combination of i and j.

Combining these, we get m * n * k iterations in total. Within the innermost loop, all operations are constant time. Hence, the time complexity is O(m * n * k).

Space Complexity:

The space complexity is determined by the size of the dp array, which stores intermediate counts for each cell and each possible remainder modulo k:

• The dp array is a 3-dimensional array with dimensions m, n, and k.
• This results in a space requirement for m * n * k integers.

Hence, the space complexity of the code is also O(m * n * k).

Learn more about how to find time and space complexity quickly using problem constraints.