What is the time complexity of Minimum Time to Visit a Cell In a Grid?

The optimal solution has a time complexity of O(m*n*log(m*n)).

What data structures are used to solve Minimum Time to Visit a Cell In a Grid?

This problem uses Array, Breadth-First Search, Graph, Heap (Priority Queue), Matrix, Shortest Path. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum Time to Visit a Cell In a Grid?

Insight 1: Dijkstra's algorithm or a variant of it is suitable for finding the shortest path (minimum time) in a graph with non-negative edge weights. The grid can be viewed as a graph where cells are nodes and moves between adjacent cells are edges. Insight 2: A priority queue is used to efficiently select the cell with the smallest time to visit next.

Which companies ask LeetCode 2577. Minimum Time to Visit a Cell In a Grid in interviews?

This problem is frequently asked at Atlassian.

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Minimum Time to Visit a Cell In a Grid - LeetCode 2577 Solution

Q: What is the space complexity of Minimum Time to Visit a Cell In a Grid?

The space complexity is O(m*n).

Minimum Time to Visit a Cell In a Grid - Complete Solution Guide

Minimum Time to Visit a Cell In a Grid is LeetCode problem 2577, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a m x n matrix grid consisting of non-negative integers where grid[row][col] represents the minimum time required to be able to visit the cell (row, col) , which means you can visit the cell (row, col) only when the time you visit it is greater than or equal to grid[row][col] . You are standing in the top-left cell of the matrix in the 0 th second, and you must move to any adjacent cell in the four directions: up, down, left, and right. Each move you make takes 1 second. Return the

Detailed Explanation

The problem asks us to find the minimum time required to reach the bottom-right cell of a grid from the top-left cell. We can only move to adjacent cells (up, down, left, right), and each move takes 1 second. Each cell in the grid has a 'minimum visit time'; we can only enter a cell if our current time is greater than or equal to that cell's value. If it's impossible to reach the bottom-right cell, return -1. The grid[0][0] is always 0, meaning we can start immediately.

Solution Approach

The provided solution uses Dijkstra's algorithm with a priority queue to find the minimum time to reach the bottom-right cell. The grid is treated as a graph. The priority queue stores tuples of (time, row, column), sorted by time. The algorithm explores the grid, updating the 'dist' array (minimum time to reach each cell) as it goes. For each neighbor, it calculates the 'tentative_time' to reach it. This time considers both the travel time and the cell's minimum visit time. Finally, it returns the minimum time to reach the destination or -1 if it's unreachable.

Step-by-Step Algorithm

Step 1: Check if the initial move is possible. If both grid[0][1] and grid[1][0] are greater than 1, it's impossible to move from grid[0][0] at time 0, so return -1.
Step 2: Initialize a 'dist' array to store the minimum time to reach each cell, setting all values to infinity except for the starting cell (0, 0), which is set to 0.
Step 3: Create a priority queue 'pq' and add the starting cell (0, 0) with time 0. The priority queue is ordered by time.
Step 4: While the priority queue is not empty:
Step 5: Extract the cell with the minimum time ('t', 'r', 'c') from the priority queue.
Step 6: If the current time 't' is greater than the current shortest time 'dist[r][c]' to reach the cell (r, c), continue to the next iteration (optimization to avoid revisiting cells with larger times).
Step 7: If the current cell is the destination (bottom-right), return the time 't'.
Step 8: Iterate through the four possible directions (up, down, left, right).
Step 9: Calculate the coordinates ('nr', 'nc') of the neighboring cell.
Step 10: If the neighboring cell is within the grid boundaries, calculate the 'tentative_time' to reach the neighbor. This is the maximum of (current time + 1) and the neighbor's 'minimum visit time' (grid[nr][nc]).
Step 11: If the difference between the 'tentative_time' and the current time 't' is even, increment 'tentative_time' by one. This waiting logic ensures the fastest possible arrival.
Step 12: If the 'new_time' (tentative_time with possible increment) is less than the current minimum time to reach the neighbor ('dist[nr][nc]'), update 'dist[nr][nc]' with the 'new_time' and add the neighbor to the priority queue with the updated time.
Step 13: If the loop finishes without reaching the destination, return -1 (no path found).

Key Insights

Insight 1: Dijkstra's algorithm or a variant of it is suitable for finding the shortest path (minimum time) in a graph with non-negative edge weights. The grid can be viewed as a graph where cells are nodes and moves between adjacent cells are edges.
Insight 2: A priority queue is used to efficiently select the cell with the smallest time to visit next.
Insight 3: The 'minimum visit time' constraint of each cell influences the overall time taken to reach a cell. We might need to wait at a cell before moving to the next if we arrive 'too early'. There's an additional waiting logic, if (tentative_time - t) % 2 == 0, we need to wait one more second.

Complexity Analysis

Time Complexity: O(m*n*log(m*n))

Space Complexity: O(m*n)

Topics

This problem involves: Array, Breadth-First Search, Graph, Heap (Priority Queue), Matrix, Shortest Path.

Companies

Asked at: Atlassian.

Minimum Time to Visit a Cell In a Grid | LeetCode 2577

Hard

Array Breadth-First Search Graph Heap (Priority Queue)Matrix Shortest Path

Time: O(m*n*log(m*n))Space: O(m*n)

Solution Code

Available in:

import java.util.PriorityQueue;
import java.util.Arrays;

class Solution {
    public int minimumTime(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;

        if (grid[0][1] > 1 && grid[1][0] > 1) {
            return -1;
        }

        int[][] dist = new int[m][n];
        for (int i = 0; i < m; i++) {
            Arrays.fill(dist[i], Integer.MAX_VALUE);
        }

        dist[0][0] = 0;

        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[0] - b[0]); // (time, r, c)
        pq.offer(new int[]{0, 0, 0});

        int[][] directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

        while (!pq.isEmpty()) {
            int[] curr = pq.poll();
            int t = curr[0];
            int r = curr[1];
            int c = curr[2];
            
            if (t > dist[r][c]) {
                continue;
            }
            
            if (r == m - 1 && c == n - 1) {
                return t;
            }

            for (int[] direction : directions) {
                int dr = direction[0];
                int dc = direction[1];

                int nr = r + dr;
                int nc = c + dc;

                if (nr >= 0 && nr < m && nc >= 0 && nc < n) {
                    int tentative_time = Math.max(t + 1, grid[nr][nc]);
                    
                    int new_time;
                    if ((tentative_time - t) % 2 == 0) {
                        new_time = tentative_time + 1;
                    } else {
                        new_time = tentative_time;
                    }

                    if (new_time < dist[nr][nc]) {
                        dist[nr][nc] = new_time;
                        pq.offer(new int[]{new_time, nr, nc});
                    }
                }
            }
        }

        return -1;
    }
}

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Problem Info

Description

Problem 2577: Minimum Time to Visit a Cell In a Grid - Detailed explanation and solution approaches.

Topics

Array Breadth-First Search Graph Heap (Priority Queue)Matrix Shortest Path

Companies

Atlassian

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2577

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

13 steps

1
Step 1: Check if the initial move is possible. If both grid[0][1] and grid[1][0] are greater than 1, it's impossible to move from grid[0][0] at time 0, so return -1.
2
Step 2: Initialize a 'dist' array to store the minimum time to reach each cell, setting all values to infinity except for the starting cell (0, 0), which is set to 0.
3
Step 3: Create a priority queue 'pq' and add the starting cell (0, 0) with time 0. The priority queue is ordered by time.
4
Step 4: While the priority queue is not empty:
5
Step 5: Extract the cell with the minimum time ('t', 'r', 'c') from the priority queue.
6
Step 6: If the current time 't' is greater than the current shortest time 'dist[r][c]' to reach the cell (r, c), continue to the next iteration (optimization to avoid revisiting cells with larger times).
7
Step 7: If the current cell is the destination (bottom-right), return the time 't'.
8
Step 8: Iterate through the four possible directions (up, down, left, right).
9
Step 9: Calculate the coordinates ('nr', 'nc') of the neighboring cell.
10
Step 10: If the neighboring cell is within the grid boundaries, calculate the 'tentative_time' to reach the neighbor. This is the maximum of (current time + 1) and the neighbor's 'minimum visit time' (grid[nr][nc]).
11
Step 11: If the difference between the 'tentative_time' and the current time 't' is even, increment 'tentative_time' by one. This waiting logic ensures the fastest possible arrival.
12
Step 12: If the 'new_time' (tentative_time with possible increment) is less than the current minimum time to reach the neighbor ('dist[nr][nc]'), update 'dist[nr][nc]' with the 'new_time' and add the neighbor to the priority queue with the updated time.
13
Step 13: If the loop finishes without reaching the destination, return -1 (no path found).

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m*n*log(m*n))

💾

Space Complexity

O(m*n)

📝

Why?

Dijkstra's algorithm with a priority queue has a time complexity of O(E log V), where E is the number of edges and V is the number of vertices. In this case, V is m*n (the number of cells in the grid), and E is approximately 4*m*n (each cell has at most 4 neighbors). Therefore, the time complexity is O(4*m*n * log(m*n)), which simplifies to O(m*n*log(m*n)). The space complexity is O(m*n) because the 'dist' array stores the minimum time for each cell, and the priority queue can potentially hold all cells in the grid in the worst case.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

A* Search

A* search is an informed search algorithm that can be used to find the shortest path. It uses a heuristic function to estimate the cost to reach the destination from a given node, guiding the search towards the goal. The heuristic can be Manhattan distance in this case.

Time: O(m*n) in best case, O(m*n*log(m*n)) in worst case depending on the heuristicSpace: O(m*n)

Trade-offs: A* can be faster than Dijkstra's if the heuristic is well-chosen and accurate, but it can have the same worst-case complexity if the heuristic is not effective. Requires careful selection of the heuristic function.

Breadth-First Search (BFS)

BFS can be used if we discretize the time steps. We can maintain a queue of cells to visit, along with the time. However, due to the minimum visit time constraint, simple BFS will not work directly without significant modifications.

Time: Potentially very high, dependent on maximum grid value O(m * n * max(grid[i][j]))Space: O(m*n * max(grid[i][j]))

Trade-offs: BFS is not efficient for this problem because we are trying to find the *minimum time*, and BFS explores all possible paths at each time step. Because of potentially large grid values, this becomes extremely slow.

✨

Key Insights

1Insight 1: Dijkstra's algorithm or a variant of it is suitable for finding the shortest path (minimum time) in a graph with non-negative edge weights. The grid can be viewed as a graph where cells are nodes and moves between adjacent cells are edges.
2Insight 2: A priority queue is used to efficiently select the cell with the smallest time to visit next.
3Insight 3: The 'minimum visit time' constraint of each cell influences the overall time taken to reach a cell. We might need to wait at a cell before moving to the next if we arrive 'too early'. There's an additional waiting logic, if (tentative_time - t) % 2 == 0, we need to wait one more second.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Graph

Nodes connected by edges. Like a social network or city map!

Edge Cases

Edge case 1: Impossible initial move (grid[0][1] > 1 && grid[1][0] > 1). The solution explicitly checks for this condition and returns -1 if it's true.
Edge case 2: Grid is unreachable, meaning there's no path from the top-left to the bottom-right. The algorithm returns -1 in this case because the destination cell is never reached, and the priority queue becomes empty.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the 'minimum visit time' constraint correctly. Failing to check if the current time is greater than or equal to the cell's value can lead to incorrect results.

✗Common mistake 2: Forgetting to consider the case where the initial move is impossible (both adjacent cells have minimum visit times greater than 1).

✗Common mistake 3: Not handling cases with an odd vs. an even number of waits between grid[r][c] and tentative_time

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Pathfinding in video games where different areas have different access requirements (e.g., levels, resources needed).

🏦Real-world scenario 2: Robot navigation in a warehouse where certain locations require a minimum time delay for safety reasons or to wait for other operations to complete.

🎮Real-world scenario 3: Network routing where nodes have minimum latency requirements for data transmission.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 28, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimum Time to Visit a Cell In a Grid"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether breadth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?