How do you solve Count Paths With the Given XOR Value on LeetCode?

Count Paths With the Given XOR Value (LeetCode #3393) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n * maxXOR) time complexity and O(m * n * maxXOR) space complexity. Key insight: Dynamic programming can significantly reduce the time complexity compared to brute force.

What is the brute force solution for Count Paths With the Given XOR Value?

The brute force approach for Count Paths With the Given XOR Value is Brute Force, with O(2^(m+n)) time complexity and O(m+n) space complexity. We can explore all possible paths from the top-left to the bottom-right corner of the grid. For each path, we compute the XOR of the values and check if it equals k. The optimal Optimal Solution approach improves this to O(m * n * maxXOR).

What algorithm or data structure is used to solve Count Paths With the Given XOR Value?

Count Paths With the Given XOR Value uses the following concepts: Array, Dynamic Programming, Bit Manipulation, Matrix. The recommended patterns to study are: Dynamic Programming, Bit Manipulation.

What are the interview tips for Count Paths With the Given XOR Value?

Always clarify the constraints and edge cases before diving into coding. Explain your thought process clearly while coding to demonstrate your understanding. Consider both time and space complexity in your solution.

What are common mistakes when solving Count Paths With the Given XOR Value?

Not considering all possible XOR values when updating the DP table. Failing to account for large numbers leading to overflow without modulo.

#3393

Count Paths With the Given XOR Value

Medium

ArrayDynamic ProgrammingBit ManipulationMatrixDynamic ProgrammingBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

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

We can use dynamic programming to store the number of ways to reach each cell with a specific XOR value. This avoids recalculating paths and significantly reduces the number of computations.

⚙️

Algorithm

3 steps

1Step 1: Create a 3D DP array where dp[i][j][x] represents the number of ways to reach cell (i, j) with XOR value x.
2Step 2: Initialize dp[0][0][grid[0][0]] to 1 since there's one way to start at the top-left corner.
3Step 3: Iterate through each cell and update the DP table based on possible moves (down and right) while maintaining the XOR value.

solution.py16 lines

1def countPaths(grid, k):
2    from collections import defaultdict
3    m, n = len(grid), len(grid[0])
4    dp = defaultdict(int)
5    dp[(0, 0, grid[0][0])] = 1
6    mod = 10**9 + 7
7    for i in range(m):
8        for j in range(n):
9            for x in list(dp.keys()):
10                if x[0] == i and x[1] == j:
11                    count = dp[x]
12                    if i + 1 < m:
13                        dp[(i + 1, j, x[2] ^ grid[i + 1][j])] += count
14                    if j + 1 < n:
15                        dp[(i, j + 1, x[2] ^ grid[i][j + 1])] += count
16    return sum(dp[(m - 1, n - 1, x)] for x in dp if x[0] == m - 1 and x[1] == n - 1 and x[2] == k) % mod

ℹ

Complexity note: The time complexity is based on the number of cells and the maximum possible XOR values we might encounter. The space complexity is due to storing results for each cell and XOR value.

1Dynamic programming can significantly reduce the time complexity compared to brute force.
2Understanding how XOR works is crucial for solving this problem effectively.

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