Problem Description

The problem poses a scenario where we have a landmass represented by a binary matrix with 1s as water and 0s as land. We start with a matrix that is all land on day 0 and each day one cell gets flooded and turns into water. The goal is to determine the last day when it's possible to traverse from the top row to the bottom row entirely walking on land cells only. You may begin your traverse from any cell on the top row and must reach any cell on the bottom row, moving only up, down, left, or right.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: The grid in the problem can be viewed as a graph where each cell represents a node. Edges exist between adjacent cells (up, down, left, right).

Is it a tree?

• No: The graph representing the grid can have loops due to multiple pathways (cells) connected, and potentially many connected components, especially considering various states of flooding over days. It is hence not a tree structure.

Is the problem related to directed acyclic graphs (DAGs)?

• No: The structure and nature of the problem revolves around a grid with undirected connectivity between the cells and does not involve acyclic properties or directiveness pertinent to DAGs.

Is the problem related to shortest paths?

• No: The goal is to find a day where it's still possible to cross from one side to another and not about finding the shortest path across.

Does the problem involve connectivity?

• Yes: The main challenge in the problem is to determine if there is a path (connectivity) that allows crossing from one side of the grid to the other, despite cells being blocked over time due to rising water.

Does the problem have small constraints?

• Given that the problem might not explicitly specify small constraints that simplify the approach like smaller grids or unique scenarios, we would not immediately opt for methods ideally used in such conditions, which may include naive or brute force approaches.

With these considerations: The question involves checking connectivity in an evolving grid, suggesting methods like DFS or BFS are suitable for exploring possible paths in the grid as it changes from day to day.

Conclusion: The flowchart suggests using DFS for this connectivity problem in an evolving (non-static) grid environment.

Intuition

To solve this problem, we observe that traversing the entire land can be interpreted as finding a path between the top and bottom that is not cut by water. Therefore, as the days progress and more cells are flooded, we must continuously monitor when such a path becomes impossible.

The intuition behind the solution is to use a technique called "Union-Find" or "Disjoint Set Union (DSU)" to dynamically track connectivities between land cells. This data structure allows us to efficiently merge land regions and check if a continuous path from top to bottom exists.

The solution iterates from the last day to the first day (reverse flood order), treating each day's cell conversion from water to land and connecting it with other land cells adjacent to it. By doing this in reverse, we can find the moment just before the land cells' connections are severed to the point that top-to-bottom traversal is no longer possible.

For efficiency, the solution introduces two virtual nodes, 'top' and 'bottom,' to represent the connections of all the top row cells and bottom row cells, respectively. This allows for an immediate check if there's a connection from any top cell to any bottom cell representing a valid path. As soon as we discover that the 'top' and 'bottom' nodes are connected, we've found the last day when the path was possible.

Learn more about Depth-First Search, Breadth-First Search, Union Find and Binary Search patterns.

Solution Approach

The solution uses the Union-Find or Disjoint Set Union (DSU) algorithm to keep track of the connectivity between cells as land gets flooded with water. Here is a step-by-step explanation:

1. Initialization: We initialize a parent list p where each cell's index has itself as a parent. We also have a grid [[False] * col for _ in range(row)] which shows the status of each cell (land or water), and two variables top and bottom as virtual nodes to represent the top and bottom rows of the matrix.

2. Union-Find: The find function is a classic part of Union-Find that recursively finds the representative (root) of the group a node belongs to. If a node's parent is itself, it's the root; otherwise, we recursively call find on its parent, implementing path compression by setting the parent directly to the root for efficiency.

3. Reverse Flooding Process: We iterate over cells in reverse order. As we're moving backward, we pretend to dry each flooded cell to see when a path last existed from top to bottom.

4. Making Connections: We mark grid[i][j] as land, and for each adjacent land cell, we union the current cell with the adjacent cells by setting the parent of the current cell's root to be the root of the adjacent cell.

5. Linking to Virtual Nodes: If a cell is in the top row, we union it with the top node. Similarly, for the bottom row cells, we union them with the bottom node.

6. Checking Connectivity: After each union, we check if the top node is connected to the bottom node by checking if they have the same parent. If yes, we found the last day where it's possible to walk from top to bottom.

7. Return the Last Day: The first day (in reverse order) when top and bottom are connected is the answer. If no such day is found, we return 0, implying that it was always possible to cross from the top to the bottom of the matrix.

In this implementation, DSU is used effectively to manage the merging of the land areas and allows us to query connectivity status between the virtual top and bottom nodes in nearly constant time due to path compression.

The given Python code demonstrates an efficient method to solve the problem using these principles. The function check is a helper function to determine whether a neighboring cell is within bounds and is land (True).

By initializing the DSU with extra two nodes for top and bottom, we avoid checking every possible path from the top to bottom row every time, reducing the complexity significantly.

Example Walkthrough

Let's consider a small example to illustrate the solution approach.

Day 0 (initial state):

0: Land
1: Water

Matrix (binary grid):
0 0
0 0

Flooded cells by day:
Day 1: (0,0)
Day 2: (1,0)

(Note: Day count starts from 1)

Problem: Identify the last day when it was possible to traverse from the top row (first row) to the bottom row (second row) on land before the given matrix is entirely flooded.

Steps:

1. Initialization:

   • Parent list p is initialized with each cell pointing to itself.
   • Grid is all False initially (all land).
   • Virtual nodes top and bottom are introduced.

2. Matrix after Day 2:

   1 0
   1 0

   There is no way to go from the top to bottom, as the leftmost column is completely water. Our goal is to find when this disconnection happened.

3. Reverse Flooding Process: We iterate from the last flooded cell to the first one:

   • Day 2 cell (1,0) dries up. Grid now looks like this:

     1 0
     0 0

     Still no top to bottom path since the first column is disconnected at the top.
   • Day 1 cell (0,0) dries up. Grid now looks like this:

     0 0
     0 0

     There is now a clear path from the top row to the bottom row.

4. Making Connections and Linking to Virtual Nodes:

   • On drying cell (0,0), we connect it to its adjacent land cells (right cell (0,1)) and to the virtual node top.
   • On drying cell (1,0), we link it to its adjacent cell (right cell (1,1)).

5. Checking Connectivity:

   • After the flooding is reversed on Day 2, we still cannot go from the top to the bottom.
   • After the flooding is reversed on Day 1, top and bottom are now connected, indicating a path is possible.

6. Return the Last Day:

   • The last day when the top and bottom are connected is Day 1. Before Day 1, there is no connectivity after the flood; therefore, the water barrier is placed on Day 1.

For larger matrices, the approach is similar, but with more connections and cells to union and find. Using the Union-Find algorithm allows us to efficiently check connectivity without manually inspecting each potential path, which would be much slower. The two virtual nodes effectively represent all possible starting and ending points, streamlining the process.

Solution Implementation

Time and Space Complexity

The provided Python code defines a function latestDayToCross that finds the latest day you can cross from the top to the bottom of a grid by stepping only on cells that are present. It implements a version of the Union-Find algorithm to keep track of connected components within the grid.

Time Complexity

The time complexity of the function is determined by iterating over the list cells in reverse and performing Union-Find operations for each cell to determine connectivity. The main components influencing the time complexity are:

1. The initial loop over cells runs in reverse, which results in O(m) complexity where m is the number of elements in cells.
2. Inside this loop, several constant-time operations are carried out, such as value assignments and boolean checks.
3. The find function is called multiple times inside the loop. The time complexity of the find method is near-constant on average, due to path compression, but strictly it’s O(log*(n)) per call, where log* is the iterated logarithm and n is the number of cells in the grid. However, in practice, for most cases this can be considered almost O(1).
4. The loop inside the main loop iterates a constant number of times (at most 4), doing constant work each time and calling the find function.

Combining all these, the total average time complexity is O(mα(n)), with m being the length of the cells array and α(n) being the Inverse Ackermann function which is very slow-growing and treated as almost O(1) for all practical purposes.

Space Complexity

The space complexity of the latestDayToCross function comprises several components:

1. The parent array p, which is a list of size n + 2 where n is the number of cells in the grid, hence O(n).
2. The grid, which stores a row x col sized grid of the cells, leading to O(n) space.
3. Constant amount of space for variables like top, bottom, i, j, and the for loop variables.

Thus, the total space complexity is O(n), where n is the product of row and col.

Learn more about how to find time and space complexity quickly using problem constraints.