How do you solve Cherry Pickup on LeetCode?

Cherry Pickup (LeetCode #741) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n²) space complexity. Key insight: Dynamic programming can significantly optimize recursive solutions by storing intermediate results.

What is the brute force solution for Cherry Pickup?

The brute force approach for Cherry Pickup is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach explores all possible paths from the starting point to the destination and back. It collects cherries along the way, but this method is inefficient as it checks every possible route. The optimal Optimal Solution approach improves this to O(n²).

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

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

What are the interview tips for Cherry Pickup?

Clearly explain your thought process and the reasoning behind your approach. Be prepared to discuss trade-offs between different approaches, especially time and space complexities. Practice explaining your code step-by-step as if teaching someone else.

What are common mistakes when solving Cherry Pickup?

Failing to handle thorn cells correctly, leading to incorrect path calculations. Not considering the backtracking step properly when calculating total cherries.

#741

Cherry Pickup

Hard

ArrayDynamic ProgrammingMatrixDynamic ProgrammingBacktrackingGrid Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n²)

💡

Intuition

Time O(n²)Space O(n²)

The optimal solution uses dynamic programming to store the maximum cherries collected at each cell during the forward and backward traversal. This avoids redundant calculations and significantly reduces the time complexity.

⚙️

Algorithm

3 steps

1Step 1: Create a DP table to store the maximum cherries collected at each cell during the forward traversal.
2Step 2: Fill the DP table by iterating through the grid, updating the maximum cherries possible at each cell.
3Step 3: After reaching the destination, reset the grid and repeat the process for the backward traversal, updating the DP table again.

solution.py16 lines

1# Full working Python code
2class Solution:
3    def cherryPickup(self, grid):
4        n = len(grid)
5        dp = [[0] * n for _ in range(n)]
6        for i in range(n):
7            for j in range(n):
8                if grid[i][j] == -1:
9                    dp[i][j] = -1
10                else:
11                    if i > 0:
12                        dp[i][j] = max(dp[i][j], dp[i-1][j])
13                    if j > 0:
14                        dp[i][j] = max(dp[i][j], dp[i][j-1])
15                    dp[i][j] += grid[i][j]
16        return max(0, dp[n-1][n-1])

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loops filling the DP table, but we now use additional space to store results, leading to O(n²) space complexity.

1Dynamic programming can significantly optimize recursive solutions by storing intermediate results.
2Understanding the grid boundaries and thorn placements is crucial for pathfinding.

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