How do you solve Cherry Pickup II on LeetCode?

Cherry Pickup II (LeetCode #1463) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(rows * cols^2) time complexity and O(rows * cols^2) space complexity. Key insight: Both robots can collect cherries independently, but they must avoid double counting when they land on the same cell.

What is the brute force solution for Cherry Pickup II?

The brute force approach for Cherry Pickup II is Brute Force, with O(3^(rows)) time complexity and O(1) space complexity. The brute-force approach involves exploring all possible paths for both robots to collect cherries. This method is simple but inefficient, as it checks every combination of movements, leading to exponential time complexity. The optimal Optimal Solution approach improves this to O(rows * cols^2).

What algorithm or data structure is used to solve Cherry Pickup II?

Cherry Pickup II uses the following concepts: Array, Dynamic Programming, Matrix. The recommended patterns to study are: Dynamic Programming, Matrix Traversal.

What are the interview tips for Cherry Pickup II?

Clearly explain your thought process as you develop the solution. Start with a brute-force approach to show your understanding before optimizing. Practice similar problems to become familiar with dynamic programming patterns.

What are common mistakes when solving Cherry Pickup II?

Failing to handle the case where both robots land on the same cell correctly. Not considering edge cases, such as when the grid has only one row or one column.

#1463

Cherry Pickup II

Hard

ArrayDynamic ProgrammingMatrixDynamic ProgrammingMatrix Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(3^(rows))	O(rows * cols^2)
Space	O(1)	O(rows * cols^2)

💡

Intuition

Time O(rows * cols^2)Space O(rows * cols^2)

The optimal solution uses dynamic programming to store results of subproblems, allowing us to avoid redundant calculations. By defining a 3D DP array, we can efficiently compute the maximum cherries collected by both robots.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP array where dp[i][j][k] represents the maximum cherries collected when Robot 1 is at column j and Robot 2 is at column k on row i.
2Step 2: Iterate through each row and update the DP array based on possible movements of both robots.
3Step 3: Return the maximum cherries collected from the last row.

solution.py16 lines

1def cherryPickup(grid):
2    rows, cols = len(grid), len(grid[0])
3    dp = [[[-1] * cols for _ in range(cols)] for _ in range(rows)]
4    for j in range(cols):
5        for k in range(cols):
6            dp[0][j][k] = grid[0][j] + (grid[0][k] if j != k else 0)
7    for i in range(1, rows):
8        for j in range(cols):
9            for k in range(cols):
10                max_cherries = 0
11                for new_j in [j-1, j, j+1]:
12                    for new_k in [k-1, k, k+1]:
13                        if 0 <= new_j < cols and 0 <= new_k < cols:
14                            max_cherries = max(max_cherries, dp[i-1][new_j][new_k])
15                dp[i][j][k] = grid[i][j] + (grid[i][k] if j != k else 0) + max_cherries
16    return max(dp[rows-1][j][k] for j in range(cols) for k in range(cols))

ℹ

Complexity note: The time complexity is polynomial because we iterate through each row and each possible combination of robot positions, leading to a manageable number of calculations.

1Both robots can collect cherries independently, but they must avoid double counting when they land on the same cell.
2Dynamic programming helps break down the problem into manageable subproblems, allowing for efficient computation.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.