How do you solve Minimum Number of Visited Cells in a Grid on LeetCode?

Minimum Number of Visited Cells in a Grid (LeetCode #2617) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Understanding how to navigate the grid based on cell values is crucial.

What is the brute force solution for Minimum Number of Visited Cells in a Grid?

The brute force approach for Minimum Number of Visited Cells in a Grid is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach explores all possible paths from the top-left to the bottom-right cell by recursively visiting each valid cell. This method guarantees finding a path if it exists but is inefficient due to its exponential time complexity. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Minimum Number of Visited Cells in a Grid?

Always consider both time and space complexity when proposing solutions. Explain your thought process clearly while coding to demonstrate your understanding. Practice dry runs of your algorithm on paper to catch potential bugs.

What are common mistakes when solving Minimum Number of Visited Cells in a Grid?

Not properly managing visited cells, which can lead to infinite loops. Overlooking edge cases where no path exists.

#2617

Minimum Number of Visited Cells in a Grid

Hard

ArrayDynamic ProgrammingStackBreadth-First SearchUnion-FindHeap (Priority Queue)MatrixMonotonic StackBreadth-First SearchPriority QueueDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

The optimal approach uses a breadth-first search (BFS) with a priority queue to efficiently explore the grid. By prioritizing cells based on their potential to reach the destination, we minimize the number of cells visited.

⚙️

Algorithm

4 steps

1Step 1: Initialize a priority queue and start from the top-left cell (0, 0) with a distance of 1.
2Step 2: For each cell, calculate the valid rightward and downward moves based on the cell's value.
3Step 3: Use the priority queue to explore cells in order of their distance, ensuring the shortest path is found first.
4Step 4: If the bottom-right cell is reached, return the count of visited cells; otherwise, return -1 if no path exists.

solution.py22 lines

1# Full working Python code
2import heapq
3
4def minVisitedCells(grid):
5    m, n = len(grid), len(grid[0])
6    pq = [(1, 0, 0)]  # (steps, row, col)
7    visited = set()
8    visited.add((0, 0))
9
10    while pq:
11        steps, i, j = heapq.heappop(pq)
12        if (i, j) == (m - 1, n - 1):
13            return steps
14        for k in range(j + 1, min(n, j + grid[i][j] + 1)):
15            if (i, k) not in visited:
16                visited.add((i, k))
17                heapq.heappush(pq, (steps + 1, i, k))
18        for k in range(i + 1, min(m, i + grid[i][j] + 1)):
19            if (k, j) not in visited:
20                visited.add((k, j))
21                heapq.heappush(pq, (steps + 1, k, j))
22    return -1

ℹ

Complexity note: The time complexity is O(n log n) due to the priority queue operations, where n is the number of cells. The space complexity is O(n) for storing visited cells and the priority queue.

1Understanding how to navigate the grid based on cell values is crucial.
2Using a priority queue can significantly optimize the search process.

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