Cherry Pickup

Hard

Google

2 views

Topics:

ArraysDynamic Programming

You are given an n x n grid representing a field of cherries. Each cell is one of three possible integers: 0 (empty), 1 (cherry), or -1 (thorn, blocking the way).

Your task is to find the maximum number of cherries you can collect by following these rules:

Start at the top-left corner (0, 0) and reach the bottom-right corner (n - 1, n - 1) by moving only right or down through valid cells (0 or 1).
After reaching (n - 1, n - 1), return to (0, 0) by moving only left or up through valid cells.
When you pass through a cell containing a cherry (1), you pick it up, and the cell becomes empty (0).
If there is no valid path between (0, 0) and (n - 1, n - 1), no cherries can be collected (return 0).

Example 1:

grid = [[0, 1, -1],
        [1, 0, -1],
        [1, 1,  1]]

Expected Output: 5

Explanation: One possible optimal path is:

Go down, down, right, right to reach (2, 2), picking up 4 cherries. The grid becomes [[0, 1, -1], [0, 0, -1], [0, 0, 0]].
Return left, up, up, left to (0, 0), picking up one more cherry.

Example 2:

grid = [[1, 1, -1],
        [1, -1, 1],
        [-1, 1, 1]]

Expected Output: 0

Constraints:

n == grid.length
n == grid[i].length
1 <= n <= 50
grid[i][j] is -1, 0, or 1.
grid[0][0] != -1
grid[n - 1][n - 1] != -1

How would you implement an algorithm to solve this problem efficiently, considering both time and space complexity? Consider the edge cases where no path exists or paths overlap.

Cherry Pickup

Hard

Google

2 views

Topics:

ArraysDynamic Programming

You are given an n x n grid representing a field of cherries. Each cell is one of three possible integers: 0 (empty), 1 (cherry), or -1 (thorn, blocking the way).

Your task is to find the maximum number of cherries you can collect by following these rules:

Start at the top-left corner (0, 0) and reach the bottom-right corner (n - 1, n - 1) by moving only right or down through valid cells (0 or 1).
After reaching (n - 1, n - 1), return to (0, 0) by moving only left or up through valid cells.
When you pass through a cell containing a cherry (1), you pick it up, and the cell becomes empty (0).
If there is no valid path between (0, 0) and (n - 1, n - 1), no cherries can be collected (return 0).

Example 1:

grid = [[0, 1, -1],
        [1, 0, -1],
        [1, 1,  1]]

Expected Output: 5

Explanation: One possible optimal path is:

Go down, down, right, right to reach (2, 2), picking up 4 cherries. The grid becomes [[0, 1, -1], [0, 0, -1], [0, 0, 0]].
Return left, up, up, left to (0, 0), picking up one more cherry.

Example 2:

grid = [[1, 1, -1],
        [1, -1, 1],
        [-1, 1, 1]]

Expected Output: 0

Constraints:

n == grid.length
n == grid[i].length
1 <= n <= 50
grid[i][j] is -1, 0, or 1.
grid[0][0] != -1
grid[n - 1][n - 1] != -1

How would you implement an algorithm to solve this problem efficiently, considering both time and space complexity? Consider the edge cases where no path exists or paths overlap.

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 are the dimensions of the grid, and what is the maximum possible size of the grid (i.e., maximum rows and columns)?
Can the cherry values in the grid be negative, zero, or only positive?
If a cell contains 0 (no cherry), can the robots still move through it?
What should I return if there is no possible path to collect any cherries?
Are the starting positions of the robots fixed at (0, 0) and (0, 0) or can they start from different cells in the top-left corner?

Brute Force Solution

Approach

Imagine two people are walking on a grid, collecting cherries. The brute force method involves trying every possible path each person can take and figuring out which combination of paths yields the most cherries.

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

Consider all possible paths for the first person to reach the destination (the bottom right corner).
For each path the first person takes, consider all possible paths for the second person to reach the destination.
For each pair of paths, figure out how many cherries were picked up. If both people are on the same square, they only pick up those cherries once.
Keep track of the maximum number of cherries picked up across all possible pairs of paths.
Once you've exhausted every possible combination of paths for both people, the maximum number of cherries you kept track of is the answer.

Code Implementation

def cherry_pickup_brute_force(grid):
    grid_size = len(grid)
    max_cherries = 0

    def find_all_paths(current_row, current_col):
        if current_row == grid_size - 1 and current_col == grid_size - 1:
            return [[(current_row, current_col)]]
        
        if current_row >= grid_size or current_col >= grid_size:
            return []
        
        paths = []
        # Explore moving down
        for path in find_all_paths(current_row + 1, current_col):
            paths.append([(current_row, current_col)] + path)
            
        # Explore moving right
        for path in find_all_paths(current_row, current_col + 1):
            paths.append([(current_row, current_col)] + path)
            
        return paths

    first_person_paths = find_all_paths(0, 0)
    second_person_paths = find_all_paths(0, 0)

    # Iterate through all combinations of paths
    for first_person_path in first_person_paths:
        for second_person_path in second_person_paths:
            cherries_collected = 0
            visited = set()

            # Calculate cherries picked by the first person
            for row, col in first_person_path:
                if (row, col) not in visited:
                    cherries_collected += grid[row][col]
                    visited.add((row, col))

            # Calculate cherries picked by the second person
            for row, col in second_person_path:
                if (row, col) not in visited:
                    cherries_collected += grid[row][col]
                    visited.add((row, col))

            #Update maximum cherries picked
            max_cherries = max(max_cherries, cherries_collected)

    return max_cherries

Big(O) Analysis

Time Complexity

O(4^(m*n)) – The brute force approach involves exploring all possible paths for two people on an m x n grid. Each person can move either down or right at each step. Therefore, each person has 2 choices at each step. For a person to reach the bottom right corner, they need to take m + n - 2 steps. Thus, the number of possible paths for one person is 2^(m+n-2), and since we have two people, we need to iterate through all combinations of their paths. This leads to a total time complexity of approximately 2^(m+n-2) * 2^(m+n-2) = 4^(m+n-2), which simplifies to O(4^(m*n)).

Space Complexity

O(1) – The described brute force approach primarily involves iterating through possible paths and calculating the cherry count for each pair of paths. It doesn't explicitly create any auxiliary data structures whose size depends on the grid size N, where N is related to the dimensions of the grid. The algorithm only needs to store a few variables to track the current path, the number of cherries picked up, and the maximum cherry count seen so far. These variables take up constant space, independent of the input size.

Optimal Solution

Approach

Imagine two people walking from the top-left to the bottom-right of a grid, collecting cherries along the way. The trick is to think of these two paths as happening simultaneously to avoid re-calculating overlaps and make the problem easier to solve.

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

Instead of thinking about two separate trips, imagine the two people walking together at the same time.
Realize that after the same number of steps, both people will be the same total distance away from the starting point, even if they took different paths.
Keep track of the maximum number of cherries you can collect at each step for both people, considering all possible paths they could have taken to get there.
If both people are at the same spot, only count the cherry at that location once.
Continue this process, moving step by step through the grid, always keeping track of the best cherry count so far for each possible position of the two people.
The final answer is the maximum number of cherries both people can have when they both reach the bottom-right corner.

Code Implementation

def cherry_pickup(grid):
    grid_size = len(grid)

    # Initialize DP table to store max cherries.
    dp_table = {}

    def get_max_cherries(row_1, column_1, row_2):
        column_2 = row_1 + column_1 - row_2

        # Base case: Out of bounds.
        if (row_1 >= grid_size or column_1 >= grid_size or
            row_2 >= grid_size or column_2 >= grid_size or
            grid[row_1][column_1] == -1 or grid[row_2][column_2] == -1):
            return float('-inf')

        # Base case: Reached the destination.
        if row_1 == grid_size - 1 and column_1 == grid_size - 1:
            return grid[row_1][column_1]

        # Memoization.
        if (row_1, column_1, row_2) in dp_table:
            return dp_table[(row_1, column_1, row_2)]

        # Calculate cherries collected.
        cherries = grid[row_1][column_1]
        if row_1 != row_2 or column_1 != column_2:
            cherries += grid[row_2][column_2]

        # Recursively explore possible paths.
        path_1 = get_max_cherries(row_1 + 1, column_1, row_2 + 1)
        path_2 = get_max_cherries(row_1 + 1, column_1, row_2)
        path_3 = get_max_cherries(row_1, column_1 + 1, row_2 + 1)
        path_4 = get_max_cherries(row_1, column_1 + 1, row_2)

        max_cherries = cherries + max(path_1, path_2, path_3, path_4)
        dp_table[(row_1, column_1, row_2)] = max_cherries

        # Store intermediate results for memoization.
        return max_cherries

    # Start both people at the top-left.
    result = get_max_cherries(0, 0, 0)

    # Handle negative results.
    return max(0, result)

Big(O) Analysis

Time Complexity

O(n^3) – The algorithm iterates through all possible steps from the top-left to the bottom-right of the grid. Since two people are moving simultaneously, and after 'step' number of moves both people have step number of moves total, we are also iterating on all possible row positions for the first person and all possible row positions for the second person. Given an n x n grid, the outer loop effectively goes from 0 to 2n. The inner loops are dependent on the outer loop, but essentially are from 0 to n. This implies a time complexity of O(n*n*n), which simplifies to O(n^3).

Space Complexity

O(N^3) – The solution utilizes dynamic programming, keeping track of the maximum cherries collected for each possible position of the two people at each step. Because the two people walk simultaneously, the DP table's dimensions depend on the step number and the row/column positions of both individuals. If we denote the grid size by N x N, where N is the number of rows and columns, the DP table will require storing results for all combinations of row positions for person 1 (row1), row positions for person 2 (row2), and the current step, which is related to the column positions. Therefore, the space needed grows proportionally to N * N * N, or N^3, due to storing the intermediate maximum cherry counts at each step for all possible positions. Thus, the auxiliary space complexity is O(N^3).

Edge Cases

Case	How to Handle
Null or empty grid	Return 0 immediately, as there are no cherries to collect.
1x1 grid	Return the value of the single cell if positive, otherwise 0.
Grid with all cells containing 0	Return 0, as no cherries can be collected.
Grid with negative values (e.g., thorns)	Handle negative values as impassable (e.g., -1) and update the DP table accordingly by setting the cherry count to a very low number.
Grid with only one path to the destination	The DP algorithm will still correctly compute the optimal solution along the single path.
Maximum sized grid (50x50) with max cherry values, potential for integer overflow	Use a data type large enough to store potentially large values in the DP table, such as long.
No path to the destination exists	Return 0, as no cherries can be collected if the destination is unreachable.
Large grid filled with mostly zero values	The algorithm might take long time traversing zeros, but optimization using memoization (DP) should avoid recomputing.