Is Sequential Grid Path Cover easy or hard?

Sequential Grid Path Cover is considered a medium difficulty problem. The traversal logic is straightforward, but recognizing when to apply bitmask state compression and memoized DFS is the main challenge.

Sequential Grid Path Cover Python/Java solution

Most implementations use recursive DFS with a bitmask to represent visited cells. The same logic translates directly across Python, Java, C++, Go, and TypeScript by mapping each grid cell to a bit index and caching DFS states.

How to solve Sequential Grid Path Cover in O(n)?

A strict O(n) solution generally does not exist because the problem explores combinations of grid paths. The optimal practical approach is DFS with state compression, which reduces repeated work using memoized bitmask states and runs in O((mn) * 2^(mn)).

What is the best approach for Sequential Grid Path Cover?

The most effective solution uses DFS with state compression. Each visited configuration is encoded as a bitmask representing which grid cells are already included in the path. The DFS state becomes (row, column, mask), and memoization prevents recomputing identical states. This reduces redundant exploration compared to plain backtracking.

Is Sequential Grid Path Cover asked at Google/Amazon/Meta?

Grid path traversal and state compression problems are common in interviews at companies like Google, Amazon, and Meta. Variants involving Hamiltonian paths, grid traversal with constraints, or bitmask DFS appear frequently in onsite algorithm rounds.

What data structure is used in Sequential Grid Path Cover?

The solution mainly relies on recursion with depth‑first search, a bitmask integer for state compression, and a memoization map or array to cache visited states. The grid itself is typically represented using a 2D array or matrix.

What is the time complexity of Sequential Grid Path Cover?

The optimized state compression DFS runs in O((mn) * 2^(mn)) time where m and n are grid dimensions. Each grid cell can appear in a bitmask state, and DFS transitions to neighboring cells. Space complexity is O(2^(mn)) for memoization of visited states.

3565. Sequential Grid Path Cover - Practice Online

You are given a 2D array grid of size m x n, and an integer k. There are k cells in grid containing the values from 1 to k exactly once, and the rest of the cells have a value 0.

You can start at any cell, and move from a cell to its neighbors (up, down, left, or right). You must find a path in grid which:

Visits each cell in grid exactly once.
Visits the cells with values from 1 to k in order.

Return a 2D array result of size (m * n) x 2, where result[i] = [x_i, y_i] represents the i^th cell visited in the path. If there are multiple such paths, you may return any one.

If no such path exists, return an empty array.

Example 1:

Input: grid = [[0,0,0],[0,1,2]], k = 2

Output: [[0,0],[1,0],[1,1],[1,2],[0,2],[0,1]]

Explanation:

Example 2:

Input: grid = [[1,0,4],[3,0,2]], k = 4

Output: []

Explanation:

There is no possible path that satisfies the conditions.

Constraints:

1 <= m == grid.length <= 5
1 <= n == grid[i].length <= 5
1 <= k <= m * n
0 <= grid[i][j] <= k
grid contains all integers between 1 and k exactly once.