Swim in Rising Water

Identifying Problem Isomorphism

“Swim in Rising Water” (#778 on LeetCode) is about finding the minimum time required to swim from the top-left cell to the bottom-right cell in a grid filled with water, where the water level in each cell rises over time.

An isomorphic problem to this is “Path With Maximum Minimum Value” (#1102 on LeetCode). In this problem, you are given a matrix of integers and you need to find the maximum minimum value among all the minimum values of each possible path from the top-left cell to the bottom-right cell.

Both problems are similar because they involve finding a path in a grid from one point to another with a certain optimization condition. In “Swim in Rising Water”, you aim to minimize the time required, which corresponds to the highest water level you encounter on your path. In “Path With Maximum Minimum Value”, you aim to maximize the minimum value among all paths, which means to find the path where the smallest value encountered is as large as possible. The commonality is that you are optimizing over the maximum of the path’s minimum value, just one in the time dimension and the other in the value dimension. Hence, both problems could potentially be solved by pathfinding algorithms like Dijkstra’s or binary search.

Both problems are quite complex as they require a good understanding of graph algorithms and pathfinding strategies. “Path With Maximum Minimum Value” is more complex due to the double optimization layer (maximizing the minimum value), which might require additional considerations during the problem-solving process.

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import heapq

class Solution:
    def swimInWater(self, grid: List[List[int]]) -> int:
        n = len(grid)
        visited = [[False]*n for _ in range(n)]
        time = [[float('inf')]*n for _ in range(n)]
        time[0][0] = grid[0][0]
        heap = [(grid[0][0], 0, 0)]
        directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]

        while heap:
            t, x, y = heapq.heappop(heap)
            if visited[x][y]:
                continue
            visited[x][y] = True
            for dx, dy in directions:
                nx, ny = x + dx, y + dy
                if 0 <= nx < n and 0 <= ny < n and max(t, grid[nx][ny]) < time[nx][ny]:
                    time[nx][ny] = max(t, grid[nx][ny])
                    heapq.heappush(heap, (time[nx][ny], nx, ny))
        return time[n-1][n-1]

Problem Classification

This problem falls under the domain of Graph Theory, specifically, it is a shortest path problem with a twist. Instead of direct distances, we are considering the time, which corresponds to the highest elevation encountered on the path.

What Components:

Grid: We are given an n x n matrix which represents a grid where each value indicates the elevation at that point.
Elevation: The values in the grid represent elevation. You can only move to a square if the elevation is at most equal to the current time.
Rainfall and Time: The rain starts to fall and at time t, the depth of the water everywhere is t.
Movement: You can swim from a square to another 4-directionally adjacent square if the elevation of both squares is at most equal to the current time.
Objective: The objective is to reach the bottom right square from the top left square, with the constraint that you can only swim to squares where the elevation is less than or equal to the current time. The aim is to return the least time needed to reach the bottom right square.

This problem can be classified as a Graph Traversal problem, more specifically a Shortest Path problem. However, unlike traditional shortest path problems, the cost of moving from one node to another isn’t fixed, but depends on the time ’t’, with higher elevations requiring more time. This puts a unique spin on it, making it a variant of the shortest path problem with a dynamic cost function. The graph is implicitly defined by the grid and the movement rules, and the vertices are the grid cells, while the edges are the possible movements between them.

Thought Process

The cues from the problem statement suggest that we need to perform a graph traversal on a 2D grid. The twist is that the ability to traverse to a neighboring node depends on the time ’t’ as well as the elevation of the current node and the neighboring node. This suggests the use of a breadth-first search (BFS) or a depth-first search (DFS), possibly augmented with a priority queue to always choose the path with the minimum time.

Here’s a step-by-step approach:

Model the problem as a graph: Each cell in the grid can be considered as a node. There is an edge from one node to another if the nodes are adjacent in the grid.
Determine the traversal order: We are asked for the minimum time, so we should traverse the nodes in increasing order of their elevations. This is because moving to a lower elevation can be done sooner. This suggests using a priority queue for the traversal.
Perform the traversal: Use a BFS or DFS to visit each node, starting from the top-left cell. The priority queue will ensure that we visit the nodes in order of their elevations. Whenever we visit a node, we update the time at which we can reach it, which is the maximum of the current time and the elevation of the node. We add all neighboring nodes to the priority queue.
Return the result: The minimum time to reach the bottom-right cell is the time associated with that cell after we have finished the traversal.

Here’s how this approach can be coded in Python3:

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import heapq

def swimInWater(grid):
    n = len(grid)
    visited = [[False]*n for _ in range(n)]
    time = [[float('inf')]*n for _ in range(n)]
    time[0][0] = grid[0][0]
    heap = [(grid[0][0], 0, 0)]
    directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]
    while heap:
        t, x, y = heapq.heappop(heap)
        if visited[x][y]:
            continue
        visited[x][y] = True
        for dx, dy in directions:
            nx, ny = x + dx, y + dy
            if 0 <= nx < n and 0 <= ny < n and max(t, grid[nx][ny]) < time[nx][ny]:
                time[nx][ny] = max(t, grid[nx][ny])
                heapq.heappush(heap, (time[nx][ny], nx, ny))
    return time[n-1][n-1]

This function starts by initializing the visited matrix to keep track of the nodes we’ve visited, and the time matrix to keep track of the minimum time needed to reach each cell. It also initializes a priority queue heap and adds the top-left cell to it. Then, it performs a BFS by repeatedly removing the cell with the lowest elevation from the heap and visiting its neighbors. When visiting a neighbor, it updates the time needed to reach it, if the new time is lower than the previously recorded time. The function continues the BFS until it has visited all cells, and finally returns the minimum time needed to reach the bottom-right cell.

The problem statement provided gives us the Solution class definition and a function signature for swimInWater(). This function takes a 2D list of integers as an input and is expected to return a single integer.

Continuing with the logic described earlier, let’s implement the code inside the swimInWater function.

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import heapq

class Solution:
    def swimInWater(self, grid: List[List[int]]) -> int:
        n = len(grid)
        visited = [[False]*n for _ in range(n)]
        time = [[float('inf')]*n for _ in range(n)]
        time[0][0] = grid[0][0]
        heap = [(grid[0][0], 0, 0)]
        directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]
        while heap:
            t, x, y = heapq.heappop(heap)
            if visited[x][y]:
                continue
            visited[x][y] = True
            for dx, dy in directions:
                nx, ny = x + dx, y + dy
                if 0 <= nx < n and 0 <= ny < n and max(t, grid[nx][ny]) < time[nx][ny]:
                    time[nx][ny] = max(t, grid[nx][ny])
                    heapq.heappush(heap, (time[nx][ny], nx, ny))
        return time[n-1][n-1]

In the code above, we:

Initialize the visited and time matrices, and the heap priority queue.
Set the initial time for the top-left cell to be its elevation.
Add the top-left cell to the priority queue.
Begin a while-loop that continues as long as the priority queue is not empty.
In each iteration of the loop, we pop a cell from the queue, check if it’s been visited before, and if so, skip the remaining steps of the loop.
If the cell hasn’t been visited before, we mark it as visited.
We then iterate over each of the cell’s neighbors. If the neighbor is within the grid bounds and we can reach it sooner than previously recorded, we update the time to reach it and add it to the priority queue.
Once the loop ends, we return the time to reach the bottom-right cell, which is the last element of the time matrix.

Language Agnostic Coding Drills

Identify distinct concepts: Here are the unique concepts present in the swimInWater function:

- **Multi-dimensional array initialization:** The solution creates two-dimensional arrays to track visited cells and the minimum time to reach each cell.

- **Use of priority queue (heap):** The solution uses a priority queue to determine the cell to visit next based on the minimum time.

- **While loop:** The solution uses a while loop to continuously process cells until no more cells are left to process.

- **Tuple unpacking:** The solution uses tuple unpacking to extract the time and coordinates of the cell to process next from the heap.

- **Control flow using continue:** The solution uses the `continue` statement to skip over already visited cells.

- **For loop over list of tuples:** The solution uses a for loop to iterate over the 4 possible directions from the current cell.

- **Coordinate manipulation and boundary checking:** The solution manipulates cell coordinates and checks whether they lie within the grid boundaries.

- **Conditional update of variables and priority queue:** The solution conditionally updates the minimum time to reach a cell and adds it to the priority queue for future processing.

List of coding concepts in increasing difficulty:

- **Multi-dimensional array initialization:** This is a basic concept in Python and most other languages, which requires understanding of how arrays work.

- **Tuple unpacking:** This is slightly more advanced than array initialization but still a fundamental concept in Python.

- **For loop over list of tuples:** This concept builds on both of the previous ones.

- **Coordinate manipulation and boundary checking:** This concept requires a basic understanding of coordinate geometry and conditional statements.

- **Control flow using continue:** This concept is more about understanding the control flow of a program, which might be tricky for beginners.

- **While loop:** While loops are generally more difficult than for loops as they require explicit handling of the termination condition.

- **Use of priority queue (heap):** Understanding how priority queues work and how to use them is more advanced and specific than the previous general programming concepts.

- **Conditional update of variables and priority queue:** This concept combines multiple previous concepts (conditional statements, priority queue, variable update) and also involves the problem-specific logic of determining when to update the variables and queue.

Problem-solving approach:

The problem is solved using a technique known as Best-First Search, which is an algorithm for traversing or searching a graph in a way that visits the most promising node chosen according to a specified rule. In this case, the rule is to always visit the cell that we can reach in the minimum time, which we keep track of using a priority queue (heap). The algorithm also keeps track of the minimum time to reach each cell, which is updated whenever we find a quicker route to a cell.

Each of the coding drills listed above contributes to implementing this algorithm. The multi-dimensional arrays store the algorithm's state, the priority queue determines the order of cell processing, the while loop encapsulates the main algorithm logic, and the for loop allows us to check all possible next cells. The coordinate manipulation and boundary checking, conditional update of variables and priority queue, and control flow using continue are all parts of the algorithm's logic that determine when and how to update the algorithm's state.

Targeted Drills in Python

1. Python Code for Identified Concepts:

Multi-dimensional array initialization:

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# Create a 3x3 array filled with zeros
n = 3
array = [[0 for _ in range(n)] for _ in range(n)]
print(array)

Tuple unpacking:

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# Assigning values to variables directly from a tuple
tuple_data = (1, 2)
a, b = tuple_data
print(a, b)

For loop over list of tuples:

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# Iterate over a list of tuples
directions = [(0, 1), (1, 0), (-1, 0), (0, -1)]
for dx, dy in directions:
    print(dx, dy)

Coordinate manipulation and boundary checking:

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# Assume x and y are coordinates of a point in a 3x3 grid
x, y = 1, 1
n = 3  # grid size
directions = [(0, 1), (1, 0), (-1, 0), (0, -1)]
for dx, dy in directions:
    nx, ny = x + dx, y + dy
    if 0 <= nx < n and 0 <= ny < n:
        print(nx, ny)

Control flow using continue:

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# Skip even numbers
for i in range(10):
    if i % 2 == 0:
        continue
    print(i)

While loop:

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# Print numbers from 0 to 9
i = 0
while i < 10:
    print(i)
    i += 1

Use of priority queue (heap):

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# Using Python's heapq module to create a min heap
import heapq
heap = []
heapq.heappush(heap, (2, 'b'))  # Push item onto heap
heapq.heappush(heap, (1, 'a'))  # Push item onto heap
print(heapq.heappop(heap))  # Pop and return smallest item from the heap

Conditional update of variables and priority queue:

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# Assume we have a variable `min_time` and a heap `pq`
import heapq
min_time = [10, 20, 30]
pq = [(10, 0), (20, 1), (30, 2)]
heapq.heapify(pq)
for i in range(3):
    time, index = pq[0]
    if time < min_time[index]:
        min_time[index] = time
        heapq.heappop(pq)
        heapq.heappush(pq, (min_time[index], index))

2. Problem-specific concepts:

One problem-specific concept here is the application of a variation of Dijkstra’s algorithm for shortest path in a grid. This concept is fundamental to our problem because we want to find the minimum time to reach the bottom right square, which is similar to finding the shortest path in a weighted graph where weights represent time to cross a cell.

3. Integrating drills to solve the problem:

The drills we identified correspond directly to different parts of the swimInWater function.

We begin by initializing our 2D arrays for visited cells and time.
We then create our priority queue with the initial cell.
Within our while loop, we first extract the cell with the minimum time from the priority queue using tuple unpacking.
We then check if we’ve already visited this cell and if so, continue to the next iteration.
Using a for loop, we iterate over all 4 possible directions, updating coordinates and checking boundaries.
If we haven’t visited a cell and it’s within boundaries, we calculate the new time to reach it.
Finally, if this new time is less than the current recorded time, we update it and add the cell to the priority queue.

This process continues until we’ve processed all cells, at which point the minimum time to reach the bottom-right cell will be in our time array.