Is Minimum Number of Visited Cells in a Grid easy or hard?

Minimum Number of Visited Cells in a Grid is classified as Hard because a straightforward BFS is too slow. The challenge is recognizing how to prune row and column searches using ordered sets, heaps, or optimized DP structures.

Minimum Number of Visited Cells in a Grid Python/Java solution

Both Python and Java implementations typically use BFS with sets or a DP approach with priority queues. Python often relies on heapq and sorted containers, while Java solutions use PriorityQueue and TreeSet for efficient range management.

How to solve Minimum Number of Visited Cells in a Grid in O(mn log n)?

Use dynamic programming with heaps that store candidate cells capable of reaching future indices. While scanning the grid, maintain row and column heaps containing the best DP values and their reachable boundaries. Remove expired entries and use the minimum valid candidate to compute the current state.

What is the best approach for Minimum Number of Visited Cells in a Grid?

The most practical approach uses BFS with row and column pruning structures such as ordered sets or heaps. It treats the grid as a graph and processes cells level by level to guarantee the shortest path. With efficient removal of visited indices, the complexity becomes about O(mn log(m+n)).

Is Minimum Number of Visited Cells in a Grid asked at Google/Amazon/Meta?

Hard grid and graph traversal problems with optimization techniques are common in interviews at companies like Google, Amazon, and Meta. This problem combines BFS, interval pruning, and priority queues, which are patterns frequently tested in senior-level algorithm interviews.

What data structure is used in Minimum Number of Visited Cells in a Grid?

Key data structures include queues for BFS traversal, ordered sets or balanced trees to track remaining indices, and priority queues (heaps) for efficient minimum queries in the DP approach. These structures prevent repeatedly scanning entire rows or columns.

What is the time complexity of Minimum Number of Visited Cells in a Grid?

The optimized solutions run in roughly O(mn log n) or O(mn log(m+n)) depending on the data structure used. Each grid cell is processed once, while heap or set operations introduce the logarithmic factor. Space complexity is O(mn) for distance tracking and auxiliary structures.

Description Hints 3 Solution Video 24:58

2617. Minimum Number of Visited Cells in a Grid

Hard24.1% acceptance3 companies

You are given a 0-indexed m x n integer matrix grid. Your initial position is at the top-left cell (0, 0).

Starting from the cell (i, j), you can move to one of the following cells:

Cells (i, k) with j < k <= grid[i][j] + j (rightward movement), or
Cells (k, j) with i < k <= grid[i][j] + i (downward movement).

Return the minimum number of cells you need to visit to reach the bottom-right cell (m - 1, n - 1). If there is no valid path, return -1.

Example 1:

Input: grid = [[3,4,2,1],[4,2,3,1],[2,1,0,0],[2,4,0,0]]
Output: 4
Explanation: The image above shows one of the paths that visits exactly 4 cells.

Example 2:

Input: grid = [[3,4,2,1],[4,2,1,1],[2,1,1,0],[3,4,1,0]]
Output: 3
Explanation: The image above shows one of the paths that visits exactly 3 cells.

Example 3:

Input: grid = [[2,1,0],[1,0,0]]
Output: -1
Explanation: It can be proven that no path exists.

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 10⁵
1 <= m * n <= 10⁵
0 <= grid[i][j] < m * n
grid[m - 1][n - 1] == 0

Asked at 3 companies

Huawei WorldQuant D. E. Shaw

Topics

Hints open one at a time

Hint 1

For each cell (i,j), it is critical to find out the minimum number of steps to reach (i,j), denoted dis[i][j], quickly, given the tight constraint.

Hint 2

Calculate dis[i][j] going left to right, top to bottom.

Hint 3

Suppose we want to calculate dis[i][j], keep track of a priority queue that stores (dis[i][k], i, k) for all k ≤ j, and another priority queue that stores (dis[k][j], k, j) for all k ≤ i.

24:58

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