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Cherry Pickup

Certification: Advanced Level Accuracy: 0% Submissions: 0 Points: 15

Write a C program to find the maximum number of cherries you can collect in a grid. You start at the top-left corner (0,0) and need to reach the bottom-right corner (n-1,n-1), then return back to the top-left corner. You can only move right or down when going to the bottom-right, and only move left or up when returning. Each cell contains a number of cherries (0 or positive integer), and once you pick cherries from a cell, it becomes 0.

Example 1
    
      
  • Input: grid = [[0,1,1,1],[1,0,1,1],[1,1,1,1],[1,1,0,0]]
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  • Output: 5
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  • Explanation: 
    • Path to bottom-right: (0,0) → (0,1) → (0,2) → (1,2) → (2,2) → (3,2) → (3,3) collecting 4 cherries. 
    • Path back to top-left: (3,3) → (2,3) → (1,3) → (0,3) → (0,2) → (0,1) → (0,0) collecting 1 more cherry. 
    • Total cherries collected = 5.
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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]]
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  • Output: 15
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  • Explanation: 
    • The optimal path collects cherries strategically on both forward and return journeys. 
    • Maximum cherries that can be collected is 15. 
    • This requires dynamic programming to find the optimal solution.
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Constraints
    
      
  • 1 ≤ grid.length ≤ 50
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  • 1 ≤ grid[i].length ≤ 50
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  • 0 ≤ grid[i][j] ≤ 100
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  • Time Complexity: O(n³)
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  • Space Complexity: O(n³)
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ArraysDynamic Programming EYLTIMindtree
Editorial

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Hint
Solution
Solution Hints
    
      
  • Think of this as two people walking simultaneously from top-left to bottom-right
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  • Use 3D dynamic programming with states (i1, j1, i2) where j2 = i1 + j1 - i2
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  • At each step, both people can move right or down
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  • If both people are at the same cell, count cherries only once
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  • Use memoization to avoid recalculating subproblems
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  • The base case is when both reach the bottom-right corner
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Steps to solve by this approach:

 Step 1: Initialize a 3D memoization array to store computed results for subproblems
 Step 2: Model the problem as two people walking simultaneously from (0,0) to (n-1,n-1)
 Step 3: Use coordinates (i1,j1) for person 1 and (i2,j2) for person 2, where j2 = i1+j1-i2
 Step 4: At each position, add cherries from both positions (count once if same position)
 Step 5: Try all four possible combinations of moves (down/right for each person)
 Step 6: Use memoization to avoid recalculating the same subproblems
 Step 7: Return the maximum cherries collected, or 0 if no valid path exists

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