To solve the Last Day Where You Can Still Cross problem, we employ binary search to determine the last possible day to cross from the top to the bottom. By simulating water flooding and checking for path availability using BFS or DFS, we can find the critical day. The binary search over valid days optimizes the search for the last day when a path still exists.
Problem Statement
You are given a matrix with rows and columns where initially, all cells are land. Each day, a new cell gets flooded with water according to a list of coordinates. The goal is to find the last day when it is still possible to walk from the top row to the bottom row, using only the land cells. You can start from any land cell in the top row and walk to any land cell in the bottom row, moving up, down, left, or right.
The challenge involves determining this last day using an efficient algorithm. A key aspect of solving this problem is to search for the valid answer space. Binary search can help narrow down the possible days, while graph traversal techniques such as BFS or DFS can validate the existence of a path across flooded cells.
Examples
Example 1
Input: row = 2, col = 2, cells = [[1,1],[2,1],[1,2],[2,2]]
Output: 2
The above image depicts how the matrix changes each day starting from day 0. The last day where it is possible to cross from top to bottom is on day 2.
Example 2
Input: row = 2, col = 2, cells = [[1,1],[1,2],[2,1],[2,2]]
Output: 1
The above image depicts how the matrix changes each day starting from day 0. The last day where it is possible to cross from top to bottom is on day 1.
Example 3
Input: row = 3, col = 3, cells = [[1,2],[2,1],[3,3],[2,2],[1,1],[1,3],[2,3],[3,2],[3,1]]
Output: 3
The above image depicts how the matrix changes each day starting from day 0. The last day where it is possible to cross from top to bottom is on day 3.
Constraints
- 2 <= row, col <= 2 * 104
- 4 <= row * col <= 2 * 104
- cells.length == row * col
- 1 <= ri <= row
- 1 <= ci <= col
- All the values of cells are unique.
Solution Approach
Binary Search on the Day
Binary search over the day values can be used to check for the last possible day when a valid path exists. By testing if a path exists on day X, we can adjust the search space until we find the last day where it's still possible to cross.
Breadth-First Search (BFS)
Once a day is chosen, BFS can be applied to explore the matrix and check if a path exists from the top to the bottom. BFS ensures that we explore all possible paths efficiently, given the constraints of the problem.
Depth-First Search (DFS)
Alternatively, DFS can be used to explore each path from top to bottom for a given day. While it may use more space compared to BFS, DFS can still be effective in validating the existence of a valid path in the flooded matrix.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The time complexity depends on the approach used, but the binary search component will reduce the number of days checked, leading to a time complexity of O(log(row * col)). For each day, BFS or DFS would take O(row * col), resulting in an overall time complexity of O(log(row * col) * (row * col)) in the worst case. Space complexity is O(row * col) due to the space needed for BFS or DFS traversal.
What Interviewers Usually Probe
- Candidate demonstrates an understanding of binary search and how to apply it to a range of possible answers.
- Candidate applies BFS/DFS to verify paths through a grid, indicating solid knowledge of graph traversal algorithms.
- Candidate identifies optimization opportunities through binary search, showing problem-solving skills in narrowing down search spaces.
Common Pitfalls or Variants
Common pitfalls
- Failing to optimize the search space by not using binary search on the valid days.
- Incorrect path validation by using inefficient graph traversal techniques or not accounting for the matrix's changing state over days.
- Overcomplicating the solution with unnecessary checks or missing edge cases like when the matrix is fully flooded.
Follow-up variants
- Use Union-Find to handle connectivity checks across the matrix during water flooding.
- Apply dynamic programming or memoization to speed up the path search process for each day.
- Solve the problem by iterating through all days and performing BFS/DFS for each day, though less optimal than binary search.
How GhostInterview Helps
- GhostInterview helps you break down the problem using binary search, providing a structured approach to narrowing the possible solution space efficiently.
- The solver assists in understanding how BFS or DFS can be applied to validate the path between the top and bottom rows as water floods the matrix.
- By focusing on graph traversal techniques in conjunction with binary search, GhostInterview ensures you understand the importance of efficient search in large grids.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
How does binary search help in solving the Last Day Where You Can Still Cross problem?
Binary search allows us to optimize the search for the last possible day to cross the matrix, reducing the number of days we need to check.
Can I use DFS instead of BFS for the Last Day Where You Can Still Cross problem?
Yes, DFS can be used instead of BFS to explore paths through the matrix. Both methods work, but BFS may be more space-efficient in this case.
What happens if I don't use binary search to narrow down the days?
Without binary search, you'd need to check every single day for a valid path, resulting in much higher time complexity.
How can I ensure my BFS or DFS is efficient for large matrices?
You can ensure efficiency by optimizing the path validation process, using visited nodes and limiting the search space to only land cells.
What are the key graph algorithms used in the Last Day Where You Can Still Cross problem?
The key algorithms are BFS or DFS for path validation and binary search for narrowing down the possible days.
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