How do you solve Minimum Time to Visit a Cell In a Grid on LeetCode?

Minimum Time to Visit a Cell In a Grid (LeetCode #2577) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(E log V) time complexity and O(m * n) space complexity. Key insight: Using a priority queue helps in exploring the least time-consuming paths first.

What is the brute force solution for Minimum Time to Visit a Cell In a Grid?

The brute force approach for Minimum Time to Visit a Cell In a Grid is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves exploring all possible paths from the top-left to the bottom-right cell. We check each path's time to see if we can reach the destination cell while satisfying the time constraints of each cell. The optimal Optimal Solution approach improves this to O(E log V).

What are the interview tips for Minimum Time to Visit a Cell In a Grid?

Explain your thought process clearly while coding. Discuss the trade-offs of different approaches before diving into code. Practice similar pathfinding problems to become familiar with priority queues.

What are common mistakes when solving Minimum Time to Visit a Cell In a Grid?

Not considering the time constraints when moving to adjacent cells. Failing to handle edge cases, such as when the destination is unreachable.

#2577

Minimum Time to Visit a Cell In a Grid

Hard

ArrayBreadth-First SearchGraph TheoryHeap (Priority Queue)MatrixShortest PathGraph TraversalDijkstra's Algorithm

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(E log V)
Space	O(1)	O(m * n)

💡

Intuition

Time O(E log V)Space O(m * n)

The optimal solution uses a priority queue (min-heap) to explore the grid in a way that always expands the least time-consuming path first. This is similar to Dijkstra's algorithm for finding the shortest path in a graph.

⚙️

Algorithm

5 steps

1Step 1: Initialize a priority queue and add the starting cell (0, 0) with time 0.
2Step 2: While the queue is not empty, extract the cell with the minimum time.
3Step 3: If this cell is the bottom-right cell, return the time.
4Step 4: For each valid adjacent cell, calculate the time needed to reach it and add it to the queue if it can be visited.
5Step 5: Keep track of visited cells to avoid cycles.

solution.py18 lines

1import heapq
2
3def minTimeToVisitCell(grid):
4    m, n = len(grid), len(grid[0])
5    pq = [(0, 0, 0)]  # (time, x, y)
6    visited = set()
7    while pq:
8        time, x, y = heapq.heappop(pq)
9        if (x, y) in visited:
10            continue
11        visited.add((x, y))
12        if x == m - 1 and y == n - 1:
13            return time
14        for dx, dy in [(1, 0), (-1, 0), (0, 1), (0, -1)]:
15            nx, ny = x + dx, y + dy
16            if 0 <= nx < m and 0 <= ny < n and time + 1 >= grid[nx][ny]:
17                heapq.heappush(pq, (max(time + 1, grid[nx][ny]), nx, ny))
18    return -1

ℹ

Complexity note: The time complexity is O(E log V) where E is the number of edges (possible moves) and V is the number of vertices (cells). The space complexity is O(m * n) due to the visited array.

1Using a priority queue helps in exploring the least time-consuming paths first.
2Tracking visited cells prevents cycles and unnecessary re-exploration.

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