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

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

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

The space complexity is O(m*n).

What data structures are used to solve Minimum Number of Visited Cells in a Grid?

This problem uses Array, Dynamic Programming, Stack, Breadth-First Search, Union Find, Heap (Priority Queue), Matrix, Monotonic Stack. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum Number of Visited Cells in a Grid?

Insight 1: Standard BFS may be inefficient as it potentially explores the same cells multiple times due to overlapping reachable areas. Avoiding revisits is crucial for optimization. Insight 2: Union Find can be used to efficiently skip already visited cells during the right and down movements, effectively reducing the search space and preventing redundant explorations.

Which companies ask LeetCode 2617. Minimum Number of Visited Cells in a Grid in interviews?

This problem is frequently asked at DE Shaw, Huawei, WorldQuant.

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Minimum Number of Visited Cells in a Grid - LeetCode 2617 Solution

Q: How to solve Minimum Number of Visited Cells in a Grid?

The solution uses a Breadth-First Search (BFS) approach combined with the Union-Find data structure to efficiently find the shortest path. The BFS explores the grid, and Union-Find is used to avoid revisiting already visited cells. For each cell, we explore the possible moves to the right and down. Union-Find is used to keep track of the visited cells in each row and column, allowing us to quickly find the next unvisited cell.

Minimum Number of Visited Cells in a Grid - Complete Solution Guide

Minimum Number of Visited Cells in a Grid is LeetCode problem 2617, 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 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]

Detailed Explanation

The problem asks us to find the minimum number of cells to visit in a grid to reach the bottom-right cell from the top-left cell. We can move either right or down from a cell (i, j). The number of steps we can take to the right is limited by `grid[i][j] + j` and downwards by `grid[i][j] + i`. If no path exists, we should return -1. The grid's bottom-right cell always has a value of 0.

Solution Approach

The solution uses a Breadth-First Search (BFS) approach combined with the Union-Find data structure to efficiently find the shortest path. The BFS explores the grid, and Union-Find is used to avoid revisiting already visited cells. For each cell, we explore the possible moves to the right and down. Union-Find is used to keep track of the visited cells in each row and column, allowing us to quickly find the next unvisited cell.

Step-by-Step Algorithm

Step 1: Initialize the `row_parents` and `col_parents` arrays for Union-Find. Each element initially points to itself, representing that all cells are unvisited.
Step 2: Initialize a queue for BFS, starting with the top-left cell (0, 0) and a distance of 1.
Step 3: Initialize a `visited` array to keep track of visited cells.
Step 4: While the queue is not empty, dequeue a cell (r, c) and its distance from the queue.
Step 5: Explore the possible rightward moves from the current cell. Use Union-Find to find the next unvisited cell in the row and mark cells visited to avoid revisiting.
Step 6: Explore the possible downward moves from the current cell. Use Union-Find to find the next unvisited cell in the column and mark cells visited to avoid revisiting.
Step 7: If the bottom-right cell is reached during either the rightward or downward movement, return the current distance + 1.
Step 8: If the queue becomes empty and the bottom-right cell has not been reached, return -1, indicating that no path exists.

Key Insights

Insight 1: Standard BFS may be inefficient as it potentially explores the same cells multiple times due to overlapping reachable areas. Avoiding revisits is crucial for optimization.
Insight 2: Union Find can be used to efficiently skip already visited cells during the right and down movements, effectively reducing the search space and preventing redundant explorations.
Insight 3: The row and column parents arrays maintain the next available cell to visit in each row and column, respectively. This significantly reduces the time complexity compared to a naive BFS approach.

Complexity Analysis

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

Space Complexity: O(m*n)

Topics

This problem involves: Array, Dynamic Programming, Stack, Breadth-First Search, Union Find, Heap (Priority Queue), Matrix, Monotonic Stack.

Companies

Asked at: DE Shaw, Huawei, WorldQuant.

Minimum Number of Visited Cells in a Grid | LeetCode 2617

Hard

Array Dynamic Programming Stack Breadth-First Search Union Find Heap (Priority Queue)Matrix Monotonic Stack

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

Solution Code

Available in:

import java.util.ArrayDeque;
import java.util.Arrays;
import java.util.Queue;

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

        if (m == 1 && n == 1) {
            return 1;
        }

        int[][] rowParents = new int[m][n + 1];
        int[][] colParents = new int[n][m + 1];

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n + 1; j++) {
                rowParents[i][j] = j;
            }
        }

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m + 1; j++) {
                colParents[i][j] = j;
            }
        }

        class UnionFind {
            public int find(int[] parent, int i) {
                ArrayDeque<Integer> pathToRoot = new ArrayDeque<>();
                int curr = i;
                while (parent[curr] != curr) {
                    pathToRoot.add(curr);
                    curr = parent[curr];
                }
                int root = curr;
                while (!pathToRoot.isEmpty()) {
                    int node = pathToRoot.removeLast();
                    parent[node] = root;
                }
                return root;
            }

            public void union(int[] parent, int i, int j) {
                int rootI = find(parent, i);
                int rootJ = find(parent, j);
                if (rootI != rootJ) {
                    parent[rootI] = rootJ;
                }
            }
        }

        UnionFind uf = new UnionFind();
        Queue<int[]> q = new ArrayDeque<>();
        q.offer(new int[]{0, 0, 1});
        boolean[][] visited = new boolean[m][n];
        for (int i = 0; i < m; i++) {
            Arrays.fill(visited[i], false);
        }
        visited[0][0] = true;

        while (!q.isEmpty()) {
            int[] curr = q.poll();
            int r = curr[0];
            int c = curr[1];
            int dist = curr[2];

            // Move right
            int limit = Math.min(n - 1, c + grid[r][c]);
            int k = uf.find(rowParents[r], c + 1);
            while (k <= limit) {
                if (!visited[r][k]) {
                    if (r == m - 1 && k == n - 1) {
                        return dist + 1;
                    }
                    visited[r][k] = true;
                    q.offer(new int[]{r, k, dist + 1});
                    uf.union(colParents[k], r, r + 1);
                }

                int nextK = uf.find(rowParents[r], k + 1);
                uf.union(rowParents[r], k, k + 1);
                k = nextK;
            }

            // Move down
            limit = Math.min(m - 1, r + grid[r][c]);
            k = uf.find(colParents[c], r + 1);
            while (k <= limit) {
                if (!visited[k][c]) {
                    if (k == m - 1 && c == n - 1) {
                        return dist + 1;
                    }
                    visited[k][c] = true;
                    q.offer(new int[]{k, c, dist + 1});
                    uf.union(rowParents[k], c, c + 1);
                }

                int nextK = uf.find(colParents[c], k + 1);
                uf.union(colParents[c], k, k + 1);
                k = nextK;
            }
        }

        return -1;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2617: Minimum Number of Visited Cells in a Grid - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Stack Breadth-First Search Union Find Heap (Priority Queue)Matrix Monotonic Stack

Companies

DE ShawHuaweiWorldQuant

Stats

Difficulty:Hard

Languages:4

Companies:3

Problem #:2617

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

8 steps

1
Step 1: Initialize the `row_parents` and `col_parents` arrays for Union-Find. Each element initially points to itself, representing that all cells are unvisited.
2
Step 2: Initialize a queue for BFS, starting with the top-left cell (0, 0) and a distance of 1.
3
Step 3: Initialize a `visited` array to keep track of visited cells.
4
Step 4: While the queue is not empty, dequeue a cell (r, c) and its distance from the queue.
5
Step 5: Explore the possible rightward moves from the current cell. Use Union-Find to find the next unvisited cell in the row and mark cells visited to avoid revisiting.
6
Step 6: Explore the possible downward moves from the current cell. Use Union-Find to find the next unvisited cell in the column and mark cells visited to avoid revisiting.
7
Step 7: If the bottom-right cell is reached during either the rightward or downward movement, return the current distance + 1.
8
Step 8: If the queue becomes empty and the bottom-right cell has not been reached, return -1, indicating that no path exists.

💡

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*alpha(m*n))

💾

Space Complexity

O(m*n)

📝

Why?

The BFS explores each cell at most once because of the `visited` array. The Union-Find operations effectively reduce the total work done because we skip large chunks of visited cells. The `row_parents` and `col_parents` data structures each occupy O(m*n) space, and the `visited` array occupies O(m*n) space as well. Therefore the total space complexity is O(m*n).

📖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

Dijkstra's Algorithm

Dijkstra's algorithm could be used to solve this problem by treating the grid as a graph where the cost of moving from one cell to another is 1. It's essentially a more general version of BFS that allows for varying edge weights. However, since all edge weights are equal (cost = 1), Dijkstra's doesn't offer any performance advantage over BFS in this specific scenario.

Time: O(E + V log V) = O(m*n + m*n log(m*n)) = O(m*n log(m*n)), where V is the number of vertices (cells) and E is the number of edges (possible moves).Space: O(m*n) for the distance array and priority queue.

Trade-offs: Dijkstra's is more complex to implement than BFS, and since the edge weights are uniform (all 1), it doesn't offer any performance benefits. BFS with Union Find provides better performance here. Dijkstra's would be beneficial only if there were varying costs to move between cells.

✨

Key Insights

1Insight 1: Standard BFS may be inefficient as it potentially explores the same cells multiple times due to overlapping reachable areas. Avoiding revisits is crucial for optimization.
2Insight 2: Union Find can be used to efficiently skip already visited cells during the right and down movements, effectively reducing the search space and preventing redundant explorations.
3Insight 3: The row and column parents arrays maintain the next available cell to visit in each row and column, respectively. This significantly reduces the time complexity compared to a naive BFS approach.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Edge Cases

Edge case 1: The grid has only one cell (m=1, n=1). The solution correctly returns 1 in this case.
Edge case 2: There is no valid path to the bottom-right cell. The solution correctly returns -1 after exploring all possible paths without reaching the destination.
Edge case 3: The grid is a rectangle, meaning 'm' is not equal to 'n'. The solution works correctly for rectangular grids since row_parents and col_parents are handled independently based on the grid dimensions.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Using a simple BFS without any optimizations can lead to TLE (Time Limit Exceeded) due to repeatedly visiting the same cells. Union-Find helps to avoid this.

✗Common mistake 2: Not handling the edge case where no path exists and returning an incorrect result. The code should return -1 when the queue becomes empty before reaching the bottom-right cell.

✗Common mistake 3: Incorrectly implementing the Union-Find data structure, leading to incorrect results or increased time complexity.

✅

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 a video game. The grid represents the game map, and the algorithm finds the shortest path for a character to move from one location to another, avoiding obstacles and previously explored areas.

🏦Real-world scenario 2: Robot navigation. A robot needs to navigate a warehouse or factory floor. The grid represents the floor plan, and the algorithm finds the most efficient route for the robot to reach its destination, considering constraints like obstacles and visited locations.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ 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: September 6, 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 14

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

@CodeNinjaJul 7

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

@TechLeadJun 30

❓

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.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

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 Number of Visited Cells 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 dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?