Question

JAVA Array and Loop Question. No DUPLICATES! I saw a similar question but it is not the same as this.

There are two arrays, one holds time and one holds amount, each corresponding to each other.

timeArray= [ 0, 2, 3, 0, 2]

amountArray= [500, 100, 400, 650, 50]

time = 4 sec

From these I must find the maximum sum of amountArray with correspondence of timeArray being the urgency #. There is a given time limit and before that time is finished I must find out the maximum amount I can get from amountArray.

So for example from the arrays above: from timeArray there are two 0s. I must pick the one with the highest amount, in this case 650. Then the next is there are two 2s, and pick the max amount which is 100, and 3 is just one. From this my max sum should be 1150. What would be the loop behind this algorithm, tried to do it but got very confused.

The trick is I can only pick one amount at a time and with each loop iteration, timeArray elements must decrease by 1. so for i =1, timeArray= [ 0, 1, 2, 1, 0], if i=2 timeArray= [0, 0, 1, 0, 0], etc... Once the time, in this case 4 iteration, is up, then it must give the max amount from the array.

It would be nice if you could help me with this. Thank you, will give thumbs up for correct code.

Answer 1

It's tricky somewhat. However, we can use "TWO LOOPS" to solve this one.

Follow the code below. Please follow the comments to better understand the program.

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SOURCE CODE

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class Program {
   public static void main (String[] args) {
       int[] timeArray= { 0, 2, 3, 0, 2};
int[] amountArray= {500, 100, 400, 650, 50};
int sum=0;
//Take the length of the array
int length=timeArray.length;
for(int i=0;i<length;i++)
{
int time=timeArray[i];
//If the time is -1, that means we have already processed that time. So, just continue with the next time.
if(time==-1)
continue;
else
{
int max=amountArray[i];
//We simpy need to see the element with same time RIGHT of "index i"
//So, take i+1
for(int j=i+1;j<length;j++)
{
//If the times are same, then make it -1,
// Just to remember that we have already seen this time
if(time==timeArray[j])
{
timeArray[j]=-1;
//Update max amount
if(max<amountArray[j])
{
max=amountArray[j];
}
}
}
//Add the maximum to the final sum
sum+=max;
System.out.println("Index="+i+" : Maximum Number= "+max);
}
}
System.out.println("The total sum is: "+sum);
   }
}

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SCREENSHOT FOR THE CODE:

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I HOPE THIS HELPS A LOT..!!

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