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Cherry Pickup

Certification: Advanced Level Accuracy: 100% Submissions: 1 Points: 15

Write a Java program to solve the Cherry Pickup problem. You are given an n x n grid representing a field of cherries, each cell is either empty (0) or contains one cherry (1). You need to collect maximum number of cherries possible by following these rules: - Start at position (0, 0) and collect cherries while moving to position (n-1, n-1) - Then return back from (n-1, n-1) to (0, 0), collecting cherries again - When you pass through a cell, you pick up all cherries, and the cell becomes empty - You can only move right or down during the first trip, and left or up during the return trip - The grid is given as a 2D array where grid[i][j] represents the number of cherries at position (i, j)

Example 1
    
  
  • Input: grid = [ 
    
    [0, 1, -1],
    
    [1, 0, -1],
    
    [1, 1, 1]
    
]
    • 
  
  • Output: 5
    • 
  
  • Explanation:

    Step 1: The grid is 3x3 with cherries (1), empty cells (0), and thorns (-1).

    Step 2: First trip path: (0,0) → (0,1) → (1,1) → (2,1) → (2,2)

    Step 3: Cherries collected: 0 + 1 + 0 + 1 + 1 = 3

    Step 4: Return trip path: (2,2) → (2,1) → (1,1) → (0,1) → (0,0)

    Step 5: Additional cherries: 0 + 0 + 0 + 0 + 0 = 0 (all already collected)

    Step 6: Max total cherries: 3 + 0 = 3

    Step 7: But there's a better path: (0,0) → (1,0) → (2,0) → (2,1) → (2,2) and back, giving 5 cherries.
    •
Example 2
    
  
  • Input: grid = [
    
    [1, 1, 1, 1, 0, 0, 0],
    
    [0, 0, 0, 1, 0, 0, 0],
    
    [0, 0, 0, 1, 0, 0, 1],
    
    [1, 0, 0, 1, 0, 0, 0],
    
    [0, 0, 0, 1, 0, 0, 0],
    
    [0, 0, 0, 1, 0, 0, 0],
    
    [0, 0, 0, 1, 1, 1, 1]

    ]
    • 
  
  • Output: 15
    • 
  
  • Explanation:

    Step 1: The grid is 7x7 with cherries (1) and empty cells (0).

    Step 2: The optimal path goes along the right border and then bottom border for the first trip.

    Step 3: Then returns following the same path, but no additional cherries are collected.

    Step 4: Total cherries collected: 15.
    •
Constraints
    
  
  • 1 <= n <= 50
    • 
  
  • grid[i][j] is 0, 1, or -1
    • 
  
  • grid[0][0] != -1
    • 
  
  • grid[n-1][n-1] != -1
    • 
  
  • Time Complexity: O(n³)
    • 
  
  • Space Complexity: O(n³)
    •
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Hint
Solution

Solution Hints

    
  
  • Suppose we are having two people starting from (0,0) and both reaching (n-1, n-1)
    • 
  
  • Both people move simultaneously, collecting cherries
    • 
  
  • If both land on the same cell, count the cherry only once
    • 
  
  • Use dynamic programming with state (row1, col1, row2) where col2 can be derived
    • 
  
  • The base case is when both reach (n-1, n-1)
    • 
  
  • For each state, try all four combinations of moves (right/down for each person)
    • 
  
  • Use memoization to avoid redundant calculations
    •

Steps to solve by this approach:

 Step 1: Realize that the problem can be reframed as having two people simultaneously traversing from (0,0) to (n-1,n-1).

 Step 2: Define a state as (r1, c1, r2), where (r1, c1) is the position of the first person and (r2, c2) is the position of the  second person. 

 Step 3: Calculate c2 using the fact that both people have traveled the same distance: r1 + c1 = r2 + c2.

 Step 4: Use dynamic programming with memoization to avoid redundant calculations for the same state.

 Step 5: For each state, try all four possible combinations of moves (right/down for each person).

 Step 6: If both people land on the same cell, count the cherry only once.

 Step 7: Return the maximum cherries that can be collected.

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