Is Minimum XOR Path in a Grid easy or hard?

The problem is generally considered medium difficulty. Computing XOR along a single path is straightforward, but minimizing the XOR across many possible paths requires careful state tracking or meet-in-the-middle optimization.

Minimum XOR Path in a Grid Python/Java solution

Python and Java implementations typically maintain a set of XOR states per cell or generate partial XOR paths using DFS. Python uses sets and dictionaries for fast lookups, while Java commonly uses HashSet or HashMap. Both approaches maintain time complexity around O(m * n * K) for DP-based solutions.

How to solve Minimum XOR Path in a Grid in O(n)?

An O(n) solution generally does not exist because the number of possible paths grows exponentially with grid dimensions. Practical solutions rely on dynamic programming over XOR states or meet-in-the-middle enumeration to reduce the exponential factor while still considering all valid paths.

What is the best approach for Minimum XOR Path in a Grid?

The most efficient strategy for larger grids is the meet-in-the-middle approach. It splits the path enumeration into two halves, computing XOR values from the start and from the destination separately, then combining them. This reduces the exponential search from O(2^(m+n)) to roughly O(2^((m+n)/2)) while still exploring all valid path XOR combinations.

Is Minimum XOR Path in a Grid asked at Google/Amazon/Meta?

XOR path and bit manipulation grid problems appear in interviews at large tech companies such as Google, Amazon, and Meta. Variants often test understanding of XOR properties, path enumeration, and optimization techniques like meet-in-the-middle or DP with state compression.

What data structure is used in Minimum XOR Path in a Grid?

Common implementations use hash sets or hash maps to store reachable XOR values for each grid position. DFS recursion stacks or iterative stacks handle traversal, while meet-in-the-middle solutions often store XOR states in maps keyed by grid coordinates.

What is the time complexity of Minimum XOR Path in a Grid?

Brute force DFS takes O(2^(m+n)) time because every right/down path is explored. A dynamic programming approach with XOR state tracking runs in about O(m * n * K), where K is the number of distinct XOR values per cell. The optimized meet-in-the-middle approach runs in roughly O(2^((m+n)/2)).

Description Hints 3 Solution Video 9:37

3882. Minimum XOR Path in a Grid

Medium43.6% acceptance1 company

You are given a 2D integer array grid of size m * n.

You start at the top-left cell (0, 0) and want to reach the bottom-right cell (m - 1, n - 1).

At each step, you may move either right or down.

The cost of a path is defined as the bitwise XOR of all the values in the cells along that path, including the start and end cells.

Return the minimum possible XOR value among all valid paths from (0, 0) to (m - 1, n - 1).

Example 1:

Input: grid = [[1,2],[3,4]]

Output: 6

Explanation:

There are two valid paths:

(0, 0) → (0, 1) → (1, 1) with XOR: 1 XOR 2 XOR 4 = 7
(0, 0) → (1, 0) → (1, 1) with XOR: 1 XOR 3 XOR 4 = 6

The minimum XOR value among all valid paths is 6.

Example 2:

Input: grid = [[6,7],[5,8]]

Output: 9

Explanation:

There are two valid paths:

(0, 0) → (0, 1) → (1, 1) with XOR: 6 XOR 7 XOR 8 = 9
(0, 0) → (1, 0) → (1, 1) with XOR: 6 XOR 5 XOR 8 = 11

The minimum XOR value among all valid paths is 9.

Example 3:

Input: grid = [[2,7,5]]

Output: 0

Explanation:

There is only one valid path:

(0, 0) → (0, 1) → (0, 2) with XOR: 2 XOR 7 XOR 5 = 0

The XOR value of this path is 0, which is the minimum possible.

Constraints:

1 <= m == grid.length <= 1000
1 <= n == grid[i].length <= 1000
m * n <= 1000
0 <= grid[i][j] <= 1023

Asked at 1 company

Google

Topics

Hints open one at a time

Hint 1

Use dynamic programming with state dp[i][j][x], where x is a possible XOR value.

Hint 2

dp[i][j][x] indicates whether it is possible to reach cell (i, j) with XOR equal to x.

Hint 3

Transition from the top left cell by updating x ^ grid[i][j], and track all reachable XOR values to find the minimum at the end cell.

9:37

Still stuck?Read 2 approaches, or watch a walkthrough.

Open solution

LanguageLoading editor…

1class Solution:
2    def minCost(self, grid: list[list[int]]) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

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

Input

[[6,7],[5,8]]

Input

[[2,7,5]]

Your code is saved on this device.Solution