/*********************************************************************
* 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[5]=false while sum[13]=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[0]=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[0]=true and 2 is in S, then sum[0+2] must also be true.
* Therefore, we set sum[0+2]=sum[2]=true. Then from sum[2]=true and
* element 4, we can say sum[2+4]=sum[6]=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[0] = 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);
}
}
