# The Partition Problemby Isai Damier, Android Engineer @ Google

```/*********************************************************************
* Author: Isai Damier
* Title: Subset Sum: Divide Candies Evenly
* Project: geekviewpoint
* Package: algorithms
*
* Statement:
*   Given n bags of candies, divide the candies between two kids
*   as evenly as possible without splitting individual bags. The candies
*   are valued by weight, in ounce (oz).
*
* Time Complexity: O(n*w)
*    where w is the total weight of all the candies together.
*
* Dynamic Programming Strategy:
*
*   The difference between this problem and the "Positive Subsequence
*   Sum [make hyperlink]" problem is that here, instead of solving for a given weight
*   x, we are ask to come up with x given the constraint that x must
*   be as close to T/2 as possible.
*
*   So let's do this. Instead of wasting too much time thinking about x,
*   let x = T/2 so that we can proceed as we did for the "Positive
*   Subsequence Sum" problem. Then after we are done filling sum{}
*   completely, we will walk the array backward until we find a positive
*   weight. For our illustration, we will use sequence S={2,4,7,9}.
*   Here are the steps:
*
*   0] Create a boolean array called sum of size x+1:
*     As you might guess, when we are done filling the array, all the
*     sub-sums between 0 and x that can be calculated from S will be
*     set to true and those that cannot be reached will be set to false.
*     For example if S={2,4,7,9} then sum=false while sum=true
*     because 4+9=13.
*
*   1] Initialize sum{} to false:
*      Before any computation is performed, assume/pretend that each
*      sub-sum is unreachable. We know that's not true, but for now
*      let's be outrageous.
*
*   2] Set sum at index 0 to true:
*     This truth is self-evident. By taking no elements from S, we end
*     up with an empty sub-sequence. Therefore we can mark sum=true,
*     since the sum of nothing is zero.
*
*   3] To fill the rest of the array, we are going to use the following
*     trick. Starting with 0, each time we find a positive sum, we will
*     add an element from S to that sum to get a greater sum. For example,
*     since sum=true and 2 is in S, then sum[0+2] must also be true.
*     Therefore, we set sum[0+2]=sum=true. Then from sum=true and
*     element 4, we can say sum[2+4]=sum=true, and so on.
*
*   4] Recall that we decided to let x=T/2. In an ideal world, we will
*    split the candies and the kids will get exactly the same amount.
*    But since that's not likely to happen in real life, here is a
*    sensible strategy. After we finish filling the sum{} array, starting
*    at x=T/2 we will check to see if sum[x] equals true. If it is, we
*    are done. If not, we decrease x by one and keep check, until we
*    find sum[x]==true.
*
**********************************************************************/
package algorithm.programming.dynamic;

public class CandyToKids {

public int[] subsetSumDivideByTwo(int[] S) {
//preliminary
int T = 0;
for (int a : S) {
T += a;
}

int x = T / 2 + 1;

boolean weight[] = new boolean[x + 1];
weight = true;
for (int a : S) {
for (int i = x; i >= a; i--) {
if (!weight[i] && weight[i - a]) {
weight[i] = true;
}
}
}
for (int i = x; i >= 0; i--) {
if (weight[i]) {
return new int[]{i, T - i};
}
}
return null;//unreachable but needed for java.
}
}
```
```package algorithm.programming.dynamic;

import org.junit.Test;
import static org.junit.Assert.*;

public class CandyToKidsTest {

/**
* Test of subsetSumDivideByTwo method, of class CandyToKids.
*/
@Test
public void testSubsetSumDivideByTwo() {
System.out.println("subsetSumDivideByTwo");

CandyToKids instance = new CandyToKids();

int[] S = {3, 5, 7, 8, 11, 13, 17, 21, 34};
int[] expResult = {60, 59};
int[] result = instance.subsetSumDivideByTwo(S);
assertArrayEquals(expResult, result);

S = new int[]{3, 5, 7, 8, 13, 17, 21, 34};
expResult = new int[]{55, 53};
result = instance.subsetSumDivideByTwo(S);
assertArrayEquals(expResult, result);

S = new int[]{11, 29, 37, 45, 59, 67, 89, 97, 37};
expResult = new int[]{234, 237};
result = instance.subsetSumDivideByTwo(S);
assertArrayEquals(expResult, result);
}
}
```