How do you solve Paths in Matrix Whose Sum Is Divisible by K on LeetCode?

Paths in Matrix Whose Sum Is Divisible by K (LeetCode #2435) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n * k) time complexity and O(m * n * k) space complexity. Key insight: The sum of the elements matters only in terms of their remainders when divided by k.

What is the time complexity of Paths in Matrix Whose Sum Is Divisible by K?

The optimal solution for Paths in Matrix Whose Sum Is Divisible by K runs in O(m * n * k) time and uses O(m * n * k) space. This complexity arises because we maintain a 3D DP array that tracks the number of paths for each cell and each possible remainder.

What is the brute force solution for Paths in Matrix Whose Sum Is Divisible by K?

The brute force approach for Paths in Matrix Whose Sum Is Divisible by K is Brute Force, with O(2^(m+n)) time complexity and O(m+n) space complexity. The brute force approach involves exploring all possible paths from the top-left to the bottom-right of the matrix. We will recursively calculate the sum of each path and check if it is divisible by k. The optimal Optimal Solution approach improves this to O(m * n * k).

What algorithm or data structure is used to solve Paths in Matrix Whose Sum Is Divisible by K?

Paths in Matrix Whose Sum Is Divisible by K uses the following concepts: Array, Dynamic Programming, Matrix. The recommended patterns to study are: Dynamic Programming, Matrix Traversal.

What are the interview tips for Paths in Matrix Whose Sum Is Divisible by K?

Clearly explain your thought process and the reasoning behind your approach. Be prepared to discuss the trade-offs between different approaches. Practice explaining your code as you write it to simulate the interview environment.

What are common mistakes when solving Paths in Matrix Whose Sum Is Divisible by K?

Forgetting to take modulo k when updating sums. Not considering edge cases like single cell matrices.

#2435

Paths in Matrix Whose Sum Is Divisible by K

Hard

ArrayDynamic ProgrammingMatrixDynamic ProgrammingMatrix Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(2^(m+n))	O(m * n * k)
Space	O(m+n)	O(m * n * k)

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Intuition

Time O(m * n * k)Space O(m * n * k)

Using dynamic programming, we can keep track of the number of ways to reach each cell with a specific sum modulo k. This way, we avoid recalculating paths and can efficiently count valid paths.

⚙️

Algorithm

3 steps

1Step 1: Create a 3D DP array where dp[i][j][r] represents the number of ways to reach cell (i, j) with a sum that gives remainder r when divided by k.
2Step 2: Initialize dp[0][0][grid[0][0] % k] = 1.
3Step 3: Iterate through each cell, updating the DP table based on the possible moves (down and right) and the current sums.

solution.py20 lines

1# Full working Python code
2MOD = 10**9 + 7
3
4def countPaths(grid, k):
5    m, n = len(grid), len(grid[0])
6    dp = [[[0] * k for _ in range(n)] for _ in range(m)]
7    dp[0][0][grid[0][0] % k] = 1
8
9    for i in range(m):
10        for j in range(n):
11            for r in range(k):
12                if dp[i][j][r] > 0:
13                    if i + 1 < m:
14                        new_r = (r + grid[i + 1][j]) % k
15                        dp[i + 1][j][new_r] = (dp[i + 1][j][new_r] + dp[i][j][r]) % MOD
16                    if j + 1 < n:
17                        new_r = (r + grid[i][j + 1]) % k
18                        dp[i][j + 1][new_r] = (dp[i][j + 1][new_r] + dp[i][j][r]) % MOD
19
20    return dp[m - 1][n - 1][0]

ℹ

Complexity note: This complexity arises because we maintain a 3D DP array that tracks the number of paths for each cell and each possible remainder.

1The sum of the elements matters only in terms of their remainders when divided by k.
2Dynamic programming allows us to efficiently count paths without redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.