# Optimum Grid Harvestby Isai Damier, Android Engineer @ Google

```/*********************************************************************
* Author: Isai Damier
* Title:  Optimum Grid Harvest
* Project: geekviewpoint
* Package: algorithms
*
* Statement:
*   Given n boxes of avocados tightly arranged in a h*w grid, and given
*   that you may only move down or to the right, amass the maximum
*
* Time-complexity: O(n)
*
* Dynamic Programming Strategy:
*
*   As it turns out this problem is easily solved by considering one
*   box of avocado at a time. For each box of avocado: look up, look left,
*   then pick the box that has more avocados and add it to the box on
*   which you are standing.
*
*   Here is how you might go about it:
*   0] Think of the boxes of avocados as an A[1..h][1..w] matrix or array.
*   1] Take a sheet of paper and draw an h*w grid on it. Say V[1..h][1..w].
*   2] In the upper-left cell, V, write the number of avocados
*      that's in the corresponding box: so V = A. This is
*      because at A there is nothing to the left or above. Simple.
*   3] Imagine you took your next step to the right, then after looking
*      up and looking left, you would have all the avocados in A
*      plus those in A. Hence, cell V would have
*      V+A avocados. Again there is if you look up.
*   4] But if instead of going right you went down, then after looking
*      up and looking left, you would have all the avocados in A
*      plus those in A. So that cell V would have
*      V+A avocados. (there is if you look left.)
*   5] After filling each new cell on your sheet of paper, you can fill
*      the next cell by evaluating the same possibilities as in step 3
*      and step 4. So your overall process is
*
*      V[i][j]=A[i][j] + max(V[i-1][j],V[i][j-1])
*
*   6] When you are done with cell V[h][w], it will have the maximum
*      number of avocados you can amass.
*
* Of course in a real implementation, unless you are told not to modify
* the input array A[1..h][1..w] you can double it as V[1..h][1..w]. In
* this implementation, we assume we should not touch the original.
*
* DETAILS YOU MAY NOT CARE FOR:
*
* You could summarize the strategy above as follows:
*
*   For each box we consider, we actually calculate the optimum solution
*   up to that box. Further, to calculate the next box, we use the
*   solutions for the previous boxes. So really we start with a small
*   grid and then keep solving for larger and larger grids until we
*   reach the final answer. Hence, we say the problem shows Optimal
*   Substructures: each smaller segment of the grid is a grid, and the
*   optimal solution to a smaller grid can be applied for solving a
*   larger grid.
*
*   Additionally, the grids overlap. What we mean is the grid
*   A[1..i][1..j] is contained in A[1..i][1..j+1] which in turn is
*   contained in A[1..i+1][1..j+1] up until A[1..h][1..w]. Hence,
*   some of the smaller grids are actually completely parts of the
*   larger grids. This trait is referred to as Overlapping Subproblems.
*
*   Consequently, although we already knew that our strategy is
*   Dynamic Programming, we can confirm that indeed it is Dynamic
*   Programming since we used a table (i.e. V[1..h][1..w]) to solve
*   a problem that exhibits both optimal substructures and overlapping
*   subproblems. You may also hear the word memoize, which basically
*   means we use V[1..h][1..w] to write down memos about each subproblem
*   we solve.
**********************************************************************/
public int gridHarvest(int[][] grid) {
if (null == grid) {
return 0;
}

int[][] value = new int[grid.length][grid.length];
int xMax = grid.length - 1;
int yMax = grid.length - 1;
for (int x = 0; x <= xMax; x++) {
for (int y = 0; y <= yMax; y++) {
amass(grid, x, y, value);
}
}
return value[xMax][yMax];
}

private void amass(int[][] grid, int x, int y, int[][] value) {
if (0 < x && 0 < y) {
value[x][y] = grid[x][y] + Math.max(value[x - 1][y], value[x][y - 1]);
} else if (0 < x) {
value[x][y] = grid[x][y] + value[x - 1][y];
} else if (0 < y) {
value[x][y] = grid[x][y] + value[x][y - 1];
} else {
value[x][y] = grid[x][y];
}
}```
```package algorithm.programming.dynamic;

import java.util.Arrays;
import java.util.List;
import org.junit.Test;
import static org.junit.Assert.*;

public class HarvestingTest {

/**
* Test of gridHarvest method, of class Harvesting.
*/
@Test
public void testGridHarvest() {
System.out.println("gridHarvest");
int[][] grid = {
{51, 42, 33, 25, 18},
{13, 21, 34, 55, 89},
{82, 63, 45, 27, 11},
{13, 17, 19, 23, 29}
};
Harvesting instance = new Harvesting();
int expResult = 344;
int result = instance.gridHarvest(grid);
assertEquals(expResult, result);
}

/**
* Test of gridHarvest method, of class Harvesting.
*/
@Test
public void testGridHarvest7() {
System.out.println("gridHarvest");
int[][] grid = {
{51, 92, 93, 97, 98},
{13, 21, 34, 55, 89},
{82, 63, 45, 27, 11},
{13, 17, 19, 23, 29}
};
Harvesting instance = new Harvesting();
int expResult = 560;
int result = instance.gridHarvest(grid);
assertEquals(expResult, result);
}

/**
* Test of gridHarvest method, of class Harvesting.
*/
@Test
public void testGridHarvestL() {
System.out.println("gridHarvest");
int[][] grid = {
{51, 42, 33, 25, 18},
{93, 21, 34, 55, 89},
{92, 63, 45, 17, 11},
{93, 67, 59, 23, 29}
};
Harvesting instance = new Harvesting();
int expResult = 507;
int result = instance.gridHarvest(grid);
assertEquals(expResult, result);
}
}```