Swim in Rising Water

Hard

Google

1 view

Topics:

Binary SearchGraphs

You are given an n x n integer matrix grid where each value grid[i][j] represents the elevation at that point (i, j). The rain starts to fall. At time t, the depth of the water everywhere is t. You can swim from a square to another 4-directionally adjacent square if and only if the elevation of both squares individually are at most t. You can swim infinite distances in zero time. Of course, you must stay within the boundaries of the grid during your swim.

Return the least time until you can reach the bottom right square (n - 1, n - 1) if you start at the top left square (0, 0).

Example 1:

Input: grid = [[0,2],[1,3]]
Output: 3
Explanation:
At time 0, you are in grid location (0, 0).
You cannot go anywhere else because 4-directionally adjacent neighbors have a higher elevation than t = 0.
You cannot reach point (1, 1) until time 3.
When the depth of water is 3, we can swim anywhere inside the grid.

Example 2:

Input: grid = [[0,1,2,3,4],[24,23,22,21,5],[12,13,14,15,16],[11,17,18,19,20],[10,9,8,7,6]]
Output: 16
Explanation: The final route is shown.
We need to wait until time 16 so that (0, 0) and (4, 4) are connected.

Constraints:

n == grid.length
n == grid[i].length
1 <= n <= 50
0 <= grid[i][j] < n<sup>2</sup>
Each value grid[i][j] is unique.

Swim in Rising Water

Hard

Google

1 view

Topics:

Binary SearchGraphs

You are given an n x n integer matrix grid where each value grid[i][j] represents the elevation at that point (i, j). The rain starts to fall. At time t, the depth of the water everywhere is t. You can swim from a square to another 4-directionally adjacent square if and only if the elevation of both squares individually are at most t. You can swim infinite distances in zero time. Of course, you must stay within the boundaries of the grid during your swim.

Return the least time until you can reach the bottom right square (n - 1, n - 1) if you start at the top left square (0, 0).

Example 1:

Input: grid = [[0,2],[1,3]]
Output: 3
Explanation:
At time 0, you are in grid location (0, 0).
You cannot go anywhere else because 4-directionally adjacent neighbors have a higher elevation than t = 0.
You cannot reach point (1, 1) until time 3.
When the depth of water is 3, we can swim anywhere inside the grid.

Example 2:

Input: grid = [[0,1,2,3,4],[24,23,22,21,5],[12,13,14,15,16],[11,17,18,19,20],[10,9,8,7,6]]
Output: 16
Explanation: The final route is shown.
We need to wait until time 16 so that (0, 0) and (4, 4) are connected.

Constraints:

n == grid.length
n == grid[i].length
1 <= n <= 50
0 <= grid[i][j] < n<sup>2</sup>
Each value grid[i][j] is unique.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the size of the grid `N`, and what are the possible values for `grid[i][j]`?
Can I assume that the grid will always be square (N x N)?
Is `grid[i][j]` guaranteed to be unique and in the range `[0, N*N - 1]`?
If there are multiple paths with the same minimum effort, is any valid path acceptable?
If it is impossible to reach the bottom right corner, what should I return?

Brute Force Solution

Approach

The goal is to find the least amount of water you need to swim through to reach the destination. The brute force method tries every possible path from the starting point to the ending point, checking how much water you'd need to swim through for each one.

Here's how the algorithm would work step-by-step:

Start at the beginning point.
Consider all possible directions you can go from your current spot.
For each direction, check if you can actually move there (staying within the boundaries).
Calculate the maximum water level you'd need to swim through up to this new spot, considering the water level of both the previous spot and the new spot.
Keep going to new possible spots from there, repeating the process of checking directions and calculating the maximum water level seen so far.
Do this until you reach the destination point. Remember to keep track of the maximum water level needed for this path.
Repeat the whole process, trying every single possible path from the starting point to the destination.
Once you've checked all possible paths, compare the maximum water levels needed for each path.
The smallest of these maximum water levels is the answer – it's the least amount of water you need to swim through to reach the destination.

Code Implementation

def swim_in_rising_water_brute_force(grid):
    grid_length = len(grid)

    def find_all_paths(row, col, current_max_elevation, path):
        # Check if we've reached the destination
        if row == grid_length - 1 and col == grid_length - 1:
            return current_max_elevation

        min_elevation = float('inf')

        # Possible directions to move: right, down, left, up
        directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]

        for delta_row, delta_col in directions:
            new_row = row + delta_row
            new_col = col + delta_col

            # Make sure the new position is within the grid bounds
            if 0 <= new_row < grid_length and 0 <= new_col < grid_length:
                # Avoid cycles by checking if the new cell is already in the current path
                if (new_row, new_col) not in path:

                    # Update max elevation seen so far
                    max_elevation_so_far = max(current_max_elevation, grid[new_row][new_col], grid[row][col])

                    new_path = path | {(new_row, new_col)}

                    elevation_for_path = find_all_paths(new_row, new_col, max_elevation_so_far, new_path)
                    min_elevation = min(min_elevation, elevation_for_path)

        return min_elevation

    # Start the search from the top-left corner
    initial_path = {(0, 0)}
    result = find_all_paths(0, 0, grid[0][0], initial_path)
    return result

Big(O) Analysis

Time Complexity

O(4^n) – The brute force solution explores all possible paths from the starting point to the ending point. In an n x n grid, at each cell, we can potentially move in four directions (up, down, left, right). Since the problem explores every path, for a grid of size n (which can be thought of as n x n), the number of possible paths can grow exponentially. Thus, each cell has 4 choices, leading to a time complexity proportional to 4 raised to the power of the number of cells visited, which can be up to the number of cells 'n' in the matrix for a path to exist, resulting in O(4^n). The total number of paths to explore dominates the runtime.

Space Complexity

O(N^2) – The brute force approach explores all possible paths, implying the need to keep track of visited locations to avoid cycles. The size of this 'visited' data structure, likely a 2D boolean array or set mirroring the input grid's dimensions, will grow proportionally to the number of cells in the grid, which is N^2 where N is the dimension of the grid. Additionally, recursion might lead to a call stack whose maximum depth could potentially visit most of the cells in the grid in the worst case. Thus, auxiliary space is dominated by the visited set, leading to O(N^2) space complexity.

Optimal Solution

Approach

The key is to find the lowest elevation we need to swim through to reach the destination. Instead of exploring all possible paths, we efficiently search for this minimum elevation using a clever method that focuses on elevation limits.

Here's how the algorithm would work step-by-step:

Imagine we're trying to find the smallest possible 'water level' that allows us to swim from the top-left to the bottom-right.
We'll use a method where we guess a water level, check if it's high enough to let us swim through, and then adjust our guess based on whether we succeeded or failed.
To check if a guessed water level is sufficient, we need to see if there's a path from start to finish where all the squares along the path are below that water level.
We can use a technique to explore the grid. If we can reach the destination using only squares below our guessed water level, then our guess was high enough.
If our water level guess was too high, we try a lower one. If it was too low, we try a higher one. We keep adjusting our guess until we find the minimum water level that works.
Each guess tells us something useful, narrowing down the possible range of water levels quickly and efficiently, so we don't have to try them all individually.

Code Implementation

def swim_in_rising_water(grid):
    grid_length = len(grid)

    def can_escape(elevation):
        visited = set()

        def depth_first_search(row, col):
            if (row < 0 or row >= grid_length or col < 0 or col >= grid_length or
                (row, col) in visited or grid[row][col] > elevation):
                return False

            if row == grid_length - 1 and col == grid_length - 1:
                return True

            visited.add((row, col))

            return (depth_first_search(row + 1, col) or
                    depth_first_search(row - 1, col) or
                    depth_first_search(row, col + 1) or
                    depth_first_search(row, col - 1))

        return depth_first_search(0, 0)

    left_elevation = grid[0][0]
    right_elevation = grid_length * grid_length
    minimum_elevation = right_elevation

    # Binary search to find the minimum elevation.
    while left_elevation <= right_elevation:
        middle_elevation = (left_elevation + right_elevation) // 2

        # If possible, try lower elevations
        if can_escape(middle_elevation):

            minimum_elevation = middle_elevation
            right_elevation = middle_elevation - 1
        else:

            # Otherwise, try higher elevations
            left_elevation = middle_elevation + 1

    return minimum_elevation

Big(O) Analysis

Time Complexity

O(n log n) – The algorithm performs a binary search on the possible water levels, which range from the minimum elevation to the maximum elevation in the grid, driving a cost of O(log n^2) which simplifies to O(log n), where n is the dimension of the grid (n x n). For each water level guess, a Depth-First Search (DFS) or Breadth-First Search (BFS) is performed on the grid of size n x n to check if a path exists from the top-left to the bottom-right, where 'n' is the length of each side of the grid. The search takes O(n^2) time in the worst case since it may visit every cell. Therefore, the overall time complexity is O(n^2 log n).

Space Complexity

O(N) – The provided plain English explanation describes exploring the grid to check if a guessed water level is sufficient, likely using a technique like Breadth-First Search (BFS) or Depth-First Search (DFS). This exploration typically involves maintaining a 'visited' set or matrix to keep track of which squares have already been explored, preventing cycles. In the worst case, this visited set or matrix could contain all N cells of the grid, where N is the number of cells in the grid (grid size * grid size). Therefore, the auxiliary space required is proportional to the input size N, leading to a space complexity of O(N).

Edge Cases

Case	How to Handle
Null or empty grid	Return 0 immediately, as no path exists.
1x1 grid	Return the single value at grid[0][0] as it's the only possible path.
Grid with all identical values	The solution should still function correctly, as the algorithm considers adjacent cells, effectively expanding outward from (0,0).
Grid with very large values causing potential integer overflow when summing costs	Ensure the data type used for cost calculation can accommodate the maximum possible sum, or use modular arithmetic if possible depending on the problem constraints.
Maximum-sized grid (e.g., 50x50 from problem statement)	Algorithm must scale efficiently (e.g., using a priority queue with Dijkstra's or A*), as a naive brute-force approach would time out.
Grid where no path exists with a cost less than infinity (unconnected regions)	If the algorithm finishes without reaching the destination, return infinity, indicating no valid path was found.
Grid with extremely skewed distribution of values (e.g., almost all 0s except for a few very large values blocking certain paths)	The priority queue should still prioritize the path with minimum 'effort' which accounts for the skewed distribution of costs by exploring minimum valued cells first.
Negative values in the grid (if the problem constraints do not explicitly forbid them)	The solution should either explicitly handle negative values by checking against them and disallowing them, or indicate the problem does not accept negative grid values.