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Fractional_KnapSack_Problem.java
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94 lines (75 loc) · 2.32 KB
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// Java program to solve fractional Knapsack Problem
import java.util.Arrays;
import java.util.Comparator;
// Greedy approach
public class FractionalKnapSack
{
// Time complexity O(n log n)
public static void main(String[] args)
{
int[] wt = {10, 40, 20, 30}; /* Here values are taken at the time of witing the code.*/
int[] val = {60, 40, 100, 120};
int capacity = 50;
double maxValue = getMaxValue(wt, val, capacity);
System.out.println("Maximum value we can obtain = " +
maxValue);
}
// function to get maximum value
private static double getMaxValue(int[] wt,
int[] val, int capacity)
{
ItemValue[] iVal = new ItemValue[wt.length];
for(int i = 0; i < wt.length; i++)
{
iVal[i] = new ItemValue(wt[i], val[i], i);
}
//sorting items by value;
Arrays.sort(iVal, new Comparator<ItemValue>()
{
@Override
public int compare(ItemValue o1, ItemValue o2)
{
return o2.cost.compareTo(o1.cost) ;
}
});
double totalValue = 0d;
for(ItemValue i: iVal)
{
int curWt = (int) i.wt;
int curVal = (int) i.val;
if (capacity - curWt >= 0)
{
// this weight can be picked while
capacity = capacity-curWt;
totalValue += curVal;
}
else
{
// item cant be picked whole
double fraction = ((double)capacity/(double)curWt);
totalValue += (curVal*fraction);
capacity = (int)(capacity - (curWt*fraction));
break;
}
}
return totalValue;
}
// item value class
static class ItemValue
{
Double cost;
double wt, val, ind;
// item value function
public ItemValue(int wt, int val, int ind)
{
this.wt = wt;
this.val = val;
this.ind = ind;
cost = new Double(val/wt );
}
}
}
/*
output for the above code:
Maximum value in Knapsack = 240
*/