Problem Description

The problem presents a scenario where you have a rectangular island bordered by two oceans—the Pacific Ocean on the left and top edges, and the Atlantic Ocean on the right and bottom edges. The island consists of a grid of cells described by an m x n matrix, where each cell has a certain height above sea level.

Rainwater can flow from higher or equal height cells to adjacent lower or equal height cells. The goal is to figure out from which cells on the island the water can flow to both oceans. This is a standard graph traversal problem where water flow is analogous to visiting nodes (cells) of the graph.

Flowchart Walkthrough

To analyze LeetCode 417. Pacific Atlantic Water Flow using the algorithm flowchart, let’s follow the decision-making steps:

Is it a graph?

• Yes: The matrix can be treated as a graph where each cell is a node, and edges exist between adjacent cells (up, down, left, right).

Is it a tree?

• No: As the graph represents cells in a matrix, it does not have a single root or a hierarchical structure typically seen in a tree.

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

• No: The problem handles flow in a matrix not governed by acyclic properties but rather by the topographical elevation constraints of each cell that determine where water can flow.

Is the problem related to shortest paths?

• No: The problem focuses on identifying cells from which water can flow to both the Pacific and Atlantic oceans, not finding the shortest path.

Does the problem involve connectivity?

• Yes: The task involves determining connectivity from cells to the borders (edges of the matrix) that represent the two oceans.

Is the graph weighted?

• No: The grid does not have weighted edges; it simply follows a rule where water flows from cells of higher or equal height to cells of lower or equal height.

Conclusion: According to the flowchart, after determining the connectivity in an unweighted graph, we realize Depth-First Search (DFS) is the suitable method to explore each cell’s connection to the ocean borders, respecting the problem constraints. Although the flowchart suggests BFS as an alternative, DFS is particularly effective here for managing to check all paths from a given cell to the ocean borders recursively.

Intuition

To solve this problem, one can use a Breadth-first search (BFS) approach from the ocean's perspective—starting from the ocean, go inland and figure out all the cells that can flow into this ocean. This is done by traversing the grid starting from the edges adjacent to the oceans and moving to cells of greater or equal height inland, as the water can only flow from high to low or equal height areas.

We use two search processes— one for each ocean. For the Pacific, we start from the top edge and the left edge of the grid because these are the sides bordering the Pacific Ocean. Similarly, for the Atlantic, we start from the bottom edge and the right edge.

As we perform BFS, we keep track of the cells visited (those that can flow into the current ocean) by using a visited set for each ocean. After completing the searches, the result will be the intersection of these two sets—that is, all the cells that can flow into both the Pacific and Atlantic oceans.

Learn more about Depth-First Search and Breadth-First Search patterns.

Solution Approach

The solution implements a Breadth-first search (BFS) algorithm. Here's a breakdown of its implementation:

• Two sets vis1 and vis2 are used to keep track of cells that can flow into the Pacific and Atlantic oceans, respectively.

• Two queues q1 and q2 are initialized to perform separate BFS for the Pacific and Atlantic oceans.

• The matrix's border cells that border the Pacific (i == 0 or j == 0) are added to q1 and marked in vis1. Analogously, the border cells that border the Atlantic (i == m - 1 or j == n - 1) are added to q2 and marked in vis2.

• The bfs function is defined, which will take a queue and a visited set as arguments. This function dequeues a cell from the queue, checks its four adjacent cells (north, south, east, west), and if an adjacent cell is within bounds, not yet visited, and its height is greater than or equal to the current cell's height, it marks this cell as visited and enqueues it.

• The bfs function is called for both q1 and vis1 and then for q2 and vis2.

• Once both BFS traversals are complete, the result is calculated as the intersection of vis1 and vis2. This is done by iterating through the entire grid and adding the cell to the result list if it's present in both visited sets.

• The algorithm uses a deque data structure from Python's collections module to perform efficient BFS.

The key algorithmic concept here is that of reverse-flow, starting from the destination (oceans) and moving upstream to discover all source points (cells that can flow into the ocean). By approaching the problem from the ocean's perspective rather than from each cell on the island, we avoid the complexity of checking each cell's individual paths to both oceans.

This solution highlights the application of graph traversal algorithms in a grid-like domain and exploits the fact that BFS efficiently explores all reachable nodes in a level-order fashion, which is essential for this type of water flow problem.

Example Walkthrough

To illustrate the solution approach let's consider a small 3x3 island matrix as an example:

Pacific ~ ~ ~
       ~ 1 2 3 ~ Atlantic
       ~ 8 9 4 ~ 
       ~ 7 6 5 ~
         ~ ~ ~ 

The numbers represent the height of each cell. Now let's walk through the Breadth-first search (BFS) algorithm:

1. We initialize two sets vis1 and vis2 for the Pacific and Atlantic oceans, respectively.

2. We also initialize two queues q1 and q2 for the BFS process.

3. Add all cells that border the Pacific Ocean (top and left edges, i == 0 or j == 0) to q1 and mark them in vis1. For our matrix, these are the cells (0,0),(0,1),(0,2),(1,0),(2,0). The Pacific queue and visited set will look as follows:

   q1 = [(0,0), (0,1), (0,2), (1,0), (2,0)]
   vis1 = {(0,0), (0,1), (0,2), (1,0), (2,0)}

4. Similarly, add all cells that border the Atlantic Ocean (bottom and right edges, i == m - 1 or j == n - 1) to q2 and mark them in vis2. For our matrix, these are the cells (0,2),(1,2),(2,2),(2,1),(2,0). The Atlantic queue and visited set will look as follows:

   q2 = [(0,2), (1,2), (2,2), (2,1), (2,0)]
   vis2 = {(0,2), (1,2), (2,2), (2,1), (2,0)}

5. Perform BFS starting from q1 and vis1. Dequeue (0,0) from q1 and explore its neighbor (1,0). Since (1,0) is already in vis1, continue dequeuing and exploring other cells. We check each dequeued cell's neighbors and if a neighbor's height is greater than or equal to the current cell and it's not already visited, we add it to the queue and to the visited set.

6. Repeat step 5 for q2 and vis2.

7. After performing BFS from both oceans, we find the intersection of vis1 and vis2 to get cells that can flow into both oceans. This results in:

   result = vis1.intersection(vis2) 
          = {(0,2), (1,2), (2,2), (2,1), (2,0)}

Thus, in this example, cells (0,2), (1,2), (2,2), (2,1), and (2,0) are the ones where water can flow to both the Pacific and Atlantic oceans. Notice that in this particular example, due to the setup and heights, all cells can flow to both oceans. However, in a more complex matrix with varying heights, only a subset of the cells would be in the final result.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the provided code is O(m * n), where m is the number of rows and n is the number of columns in the heights matrix. Here's the breakdown:

• The initialization and filling of vis1 and vis2 sets, as well as q1 and q2 queues, involve iterating over all the border cells, which in the worst case involves 2m + 2n - 4 operations, since there are m + n - 2 cells on each border. However, this number is still bounded by the size of the matrix and therefore is considered constant relative to the total number of operations.

• The main computational work happens in the two bfs calls. Each call to bfs function processes every cell at most once, because once a cell is visited, it's added to the corresponding visited set (vis1 or vis2) and will not be processed again. Therefore, each BFS call has a complexity of at most O(m * n).

• The final list comprehension combines the cells visited by both BFS calls, which, in the worst case, checks each cell in the matrix once. So, this is again O(m * n).

Since all these steps are sequential, the overall time complexity remains O(m * n).

Space Complexity

The space complexity of the code is also O(m * n):

• We store the visited states for both the Pacific and Atlantic oceans in vis1 and vis2, each of which may hold at most m * n elements if all the cells are reachable from both oceans.

• Each queue q1 and q2 could in the worst case hold the border cells which are 2(m + n - 2) as discussed earlier, but since this is bounded by O(m + n), it is smaller than the space needed for the vis1 and vis2 sets.

• Temporarily, for BFS, we may also have a queue that stores neighbors to visit. In the worst case, this can contain a number of elements equal to the entire grid (if the higher elevation cells traverse the entire grid), adding to O(m * n).

• The output list in the last list comprehension can hold at most m * n coordinates as well, however, the space for the result does not contribute to the space complexity as it is required for the output, not the computation process.

Taking all the above considerations into account, the space needed for the BFS visits and the storing of visited cells results in the total space complexity of O(m * n).

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