What is the time complexity of Last Day Where You Can Still Cross?

The optimal solution has a time complexity of O(R * C * alpha(R * C)).

What is the space complexity of Last Day Where You Can Still Cross?

The space complexity is O(R * C).

What data structures are used to solve Last Day Where You Can Still Cross?

This problem uses Array, Binary Search, Depth-First Search, Breadth-First Search, Union Find, Matrix. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Last Day Where You Can Still Cross?

Insight 1: The problem can be solved efficiently by working backward from the last day. This allows us to treat the problem as finding the *first* day on which crossing is *impossible*. Insight 2: Union Find is a suitable data structure to keep track of connected land cells. It allows us to efficiently determine if the top row and bottom row are connected by land.

Which companies ask LeetCode 1970. Last Day Where You Can Still Cross in interviews?

This problem is frequently asked at Atlassian.

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Last Day Where You Can Still Cross - LeetCode 1970 Solution

Last Day Where You Can Still Cross - Complete Solution Guide

Last Day Where You Can Still Cross is LeetCode problem 1970, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in Python3, Java, C++, C.

Problem Statement

There is a 1-based binary matrix where 0 represents land and 1 represents water. You are given integers row and col representing the number of rows and columns in the matrix, respectively. Initially on day 0 , the entire matrix is land . However, each day a new cell becomes flooded with water . You are given a 1-based 2D array cells , where cells[i] = [r i , c i ] represents that on the i th day, the cell on the r i th row and c i th column ( 1-based coordinates) will be covered with water (i.e.

Detailed Explanation

The problem describes a grid that initially consists entirely of land (represented by 0). Each day, a specific cell turns into water (represented by 1). The goal is to find the last day on which it is still possible to traverse from the top row to the bottom row by only stepping on land cells, moving only up, down, left, or right.

Solution Approach

The solution employs a Union Find data structure and iterates backward through the 'cells' array. For each day, it converts the cell to land and connects it to its adjacent land cells using Union Find. It also connects the top row cells to a virtual 'top' node and the bottom row cells to a virtual 'bottom' node. The algorithm stops when the 'top' and 'bottom' nodes become connected, indicating the first day crossing is impossible. The previous day is therefore the last day a crossing was possible.

Step-by-Step Algorithm

Step 1: Initialize a Union Find data structure with `row * col + 2` nodes. The first `row * col` nodes represent the grid cells, one node represents the virtual 'top', and one node represents the virtual 'bottom'.
Step 2: Initialize a grid with all cells as water (0).
Step 3: Iterate through the 'cells' array in reverse order (from the last day to the first).
Step 4: For each day, mark the corresponding cell in the grid as land (1).
Step 5: Connect the current land cell to its adjacent land cells (up, down, left, right) using the Union Find 'union' operation.
Step 6: If the current land cell is in the first row, connect it to the virtual 'top' node.
Step 7: If the current land cell is in the last row, connect it to the virtual 'bottom' node.
Step 8: Check if the virtual 'top' and 'bottom' nodes are connected using the Union Find 'find' operation. If they are connected, it means a path exists from the top row to the bottom row, and the current day is the first day crossing is impossible. Return the index `i` of the current day.
Step 9: If the loop finishes without finding a day where the 'top' and 'bottom' nodes are connected, return 0 (this handles cases where crossing is never possible, even on day 0).

Key Insights

Insight 1: The problem can be solved efficiently by working backward from the last day. This allows us to treat the problem as finding the *first* day on which crossing is *impossible*.
Insight 2: Union Find is a suitable data structure to keep track of connected land cells. It allows us to efficiently determine if the top row and bottom row are connected by land.
Insight 3: Adding virtual top and bottom nodes simplifies the connectivity check. We connect all cells in the first row to the top node and all cells in the last row to the bottom node. A path from the top to the bottom exists if the top and bottom nodes are in the same connected component.

Complexity Analysis

Time Complexity: O(R * C * alpha(R * C))

Space Complexity: O(R * C)

Topics

This problem involves: Array, Binary Search, Depth-First Search, Breadth-First Search, Union Find, Matrix.

Companies

Asked at: Atlassian.

Last Day Where You Can Still Cross | LeetCode 1970

Hard

Array Binary Search Depth-First Search Breadth-First Search Union Find Matrix

Time: O(R * C * alpha(R * C))Space: O(R * C)

Solution Code

Available in:

class Solution {
    private int[] parent;

    public int latestDayToCross(int row, int col, int[][] cells) {
        int n = row * col;
        parent = new int[n + 2];
        int[] size = new int[n + 2];
        
        for (int i = 0; i < n + 2; i++) {
            parent[i] = i;
            size[i] = 1;
        }

        int topSource = n;
        int bottomSink = n + 1;
        int[][] grid = new int[row][col];
        int[][] directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

        // Work backwards from the last day
        for (int i = cells.length - 1; i >= 0; i--) {
            int r = cells[i][0] - 1;
            int c = cells[i][1] - 1;

            grid[r][c] = 1; // Mark as land
            int idx = r * col + c;

            for (int[] dir : directions) {
                int nr = r + dir[0];
                int nc = c + dir[1];

                if (nr >= 0 && nr < row && nc >= 0 && nc < col && grid[nr][nc] == 1) {
                    union(idx, nr * col + nc, size);
                }
            }

            if (r == 0) union(idx, topSource, size);
            if (r == row - 1) union(idx, bottomSink, size);

            if (find(topSource) == find(bottomSink)) {
                return i;
            }
        }

        return 0;
    }

    private int find(int i) {
        if (parent[i] == i) return i;
        return parent[i] = find(parent[i]); // Path compression
    }

    private void union(int i, int j, int[] size) {
        int rootI = find(i);
        int rootJ = find(j);
        if (rootI != rootJ) {
            // Union by size
            if (size[rootI] < size[rootJ]) {
                parent[rootI] = rootJ;
                size[rootJ] += size[rootI];
            } else {
                parent[rootJ] = rootI;
                size[rootI] += size[rootJ];
            }
        }
    }
}

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Problem Info

Description

Problem 1970: Last Day Where You Can Still Cross - Detailed explanation and solution approaches.

Topics

Array Binary Search Depth-First Search Breadth-First Search Union Find Matrix

Companies

Atlassian

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1970

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

9 steps

1
Step 1: Initialize a Union Find data structure with `row * col + 2` nodes. The first `row * col` nodes represent the grid cells, one node represents the virtual 'top', and one node represents the virtual 'bottom'.
2
Step 2: Initialize a grid with all cells as water (0).
3
Step 3: Iterate through the 'cells' array in reverse order (from the last day to the first).
4
Step 4: For each day, mark the corresponding cell in the grid as land (1).
5
Step 5: Connect the current land cell to its adjacent land cells (up, down, left, right) using the Union Find 'union' operation.
6
Step 6: If the current land cell is in the first row, connect it to the virtual 'top' node.
7
Step 7: If the current land cell is in the last row, connect it to the virtual 'bottom' node.
8
Step 8: Check if the virtual 'top' and 'bottom' nodes are connected using the Union Find 'find' operation. If they are connected, it means a path exists from the top row to the bottom row, and the current day is the first day crossing is impossible. Return the index `i` of the current day.
9
Step 9: If the loop finishes without finding a day where the 'top' and 'bottom' nodes are connected, return 0 (this handles cases where crossing is never possible, even on day 0).

💡

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(R * C * alpha(R * C))

💾

Space Complexity

O(R * C)

📝

Why?

The time complexity is dominated by the iteration through all cells in reverse order (O(R*C)). The Union-Find operations (find and union) take almost constant time (O(α(R*C)) due to path compression and union by size (or rank). The space complexity is determined by the size of the `parent`, `size`, and `grid` arrays which all scale linearly with the number of cells.

📖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

Binary Search

Binary search on the number of days to find the last day a path exists. For each day being tested, create a grid and check for a path using DFS or BFS. Requires constructing the grid for each binary search iteration.

Time: O(R * C * log(R * C))Space: O(R * C)

Trade-offs: Binary Search approach is less efficient than the backward iteration with Union Find due to the overhead of constructing the grid for each binary search step and the DFS/BFS traversals within each step. Union Find is more efficient because it only processes each cell once during the backward iteration.

✨

Key Insights

1Insight 1: The problem can be solved efficiently by working backward from the last day. This allows us to treat the problem as finding the *first* day on which crossing is *impossible*.
2Insight 2: Union Find is a suitable data structure to keep track of connected land cells. It allows us to efficiently determine if the top row and bottom row are connected by land.
3Insight 3: Adding virtual top and bottom nodes simplifies the connectivity check. We connect all cells in the first row to the top node and all cells in the last row to the bottom node. A path from the top to the bottom exists if the top and bottom nodes are in the same connected component.

📚

Key Terms Explained

For Students

Array

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

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: A situation where it's impossible to cross even at day 0. The code correctly returns 0 in this scenario because the loop completes without connecting the 'top' and 'bottom' nodes.
Edge case 2: A situation where the input 'cells' might try to flood the same cell multiple times, this won't happen due to the problem constraints that 'All the values of cells are unique'.
Edge case 3: Empty matrix. The constraints ensure this is not possible (row >= 2, col >= 2).
Edge case 4: Very large matrices. While the complexity is O(R*C*alpha(R*C)), the inverse Ackermann function grows very slowly, making the time complexity effectively linear for practical input sizes.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly initializing the Union Find data structure, particularly the parent and size arrays. Failing to properly initialize these leads to incorrect connectivity results.

✗Common mistake 2: Incorrectly implementing the 'find' and 'union' operations in Union Find, such as neglecting path compression or union by rank/size. This can significantly degrade the performance of the algorithm, potentially leading to Time Limit Exceeded errors.

✗Common mistake 3: Iterating through the 'cells' array in the wrong order (forward instead of backward). This would make the problem much harder to solve.

✗Common mistake 4: Forgetting to add the virtual 'top' and 'bottom' nodes or failing to connect them to the corresponding cells in the first and last rows. This would result in incorrect connectivity checks.

✗Common mistake 5: Not accounting for 1-based indexing of the 'cells' array. The provided code correctly subtracts 1 to convert to 0-based indexing when accessing the grid.

✅

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: Network connectivity analysis: Determining the last point at which a network remains connected despite node failures (cells becoming water).

🏦Real-world scenario 2: Image segmentation: Identifying connected regions in an image up to a certain threshold (days until flooding).

🎮Real-world scenario 3: Flood fill algorithms: Simulating the spread of water in a terrain and determining when a path between two points is blocked.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

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

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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: November 18, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Last Day Where You Can Still Cross"?

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 binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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