I am trying to add a random number to a set. If it is already in the set, then the loop should continue and try again.
Which is better practice,
do {
nextChosenInt = rand.nextInt(48) + 1;
addFailed = !chosenInts.add(nextChosenInt);
}
while (addFailed);
or
do {
nextChosenInt = rand.nextInt(48) + 1;
}
while (!chosenInts.add(nextChosenInt));
I would vote for the second option here, with a slight change.
You are utilising the fact that the .add method returns a boolean which is a beneficial, in that you don't need to have a redundant flag variable which, from the context of your code, serves no other purpose than just terminating the loop.
Personally however I would choose to extract:
!chosenInts.add(nextChosenInt)
into its own method which is more descriptive and readable. By doing this, you are allowing anyone to understand this condition with no prior knowledge of the Collections API.
According to Java documentation, "if this set already contains the element, the call leaves the set unchanged and returns false." Regardless of when you add to the set, if the set already contains the element, it would remain unchanged. Thus, it doesn't really make a difference in performance.
Neither since sets can't hold duplicates. Just do it until set has a desired number of elements. But you could also do something like this.
Random rand = new Random();
int next = 50;
for (int i = 0; i < 10; i++) {
next = rand.nextInt(48) + next+1;
set.add(next);
}
They are all unique the first time (but not entirely random). To add a number that is not in the set, the following is possible.
int size = set.size();
while (set.size() == size) {
set.add(rand.nextInt(48));
}
Related
If have a workflow that removes elements of a List by a certain criteria. However certain items are skipped? Why is this happening?
List<Integer> listWithAge = new ArrayList<>();
int randomNumber = 100;
for (int i = 0; i < randomNumber; i++) {
listWithAge.add(i);
}
// this is my loop
for (int i = 0; i < listWithAge.size(); i++) {
System.out.println(i);
if ((listWithAge.get(i) % 3) == 2) listWithAge.remove(i);
}
Above code is my loop. I replaced my condition with something simpler. If I run this code my second loop only runs for 67 turns instead of 100.
It is problematic to iterate over a list and remove elements while iterating over it.
If you think about how the computer has to reconcile it, it makes sense...
Here's a thought experiment for you to go through.
If you have a list that is size 10 and you want to remove elements 1, 5, and 9 then you would think maybe the following would work:
List<String> listOfThings = ...some list with 10 things in it...;
list.remove(0);
list.remove(4);
list.remove(8);
However, after the first remove command, the list is only size 9.. Then after the second command, it's size has become 8. At this point, it hardly even makes sense to do list.remove(8) anymore because you're looking at an 8-element list and the largest index is 7.
You can also see now that the 2nd command didn't even remove the element now that you wanted.
If you want to keep this style of "remove as I go" syntax, the more appropriate way is to use Iterators. Here's an SO that talks about it and shows you the syntax you would need (see the question). It's easy to read up on elsewhere too.
How Iterator's remove method actually remove an object
Skipping a value would be the result of your list getting out of sync with your loop index because the list is reduced in size. This causes you to hop over some locations since the reduction in size affects future locations that have not been reached.
So the first thing you could do is simply correct the synchronization by decrementing i when you remove a value from the list. This will keep index at the same spot as the list shifts "left" caused by the removal.
for (int i = 0; i < listWithAge.size(); i++) {
if ((listWithAge.get(i) % 3) == 2) listWithAge.remove(i--);
}
The other option is to loop thru the list backwards.
for (int i = listWithAge.size()-1; i >= 0; i--) {
if ((listWithAge.get(i) % 3) == 2) {
listWithAge.remove(i);
}
}
This way, no values should be skipped since the removing of the element does affect the loop index's future positions relative to the changing size of the list.
But the best way would be to use an iterator as has already been mentioned by
Atmas
As a side note, I recommend you always use blocks {} even for single statements as I did above in the if block. It will save you some serious debugging time in the future when you decide you need to add additional statements and then wonder why things are no longer working.
And deleting like this from a list is very expensive, especially for large lists. I would suggest that if you don't have duplicate values, you use a Set. Otherwise, instead of deleting matching values, add the non-matching to a second list.
List<Integer> listWithAge = new ArrayList<>();
int randomNumber = 100;
for (int i = 0; i < randomNumber; i++) {
listWithAge.add(i);
}
// this is my loop
List<Integer> itemsToBeDeleted = new ArrayList<>();
for (int i = 0; i < listWithAge.size(); i++) {
System.out.println(i);
if ((listWithAge.get(i) % 3) == 2) {
itemsToBeDeleted.add(i);
}
//delete all outside the loop
//deleting inside the loop messes the indexing of the array
listWithAge.removeAll(itemsToBeDeleted);
I am passing some parameters in the URL and then I add them in a list. My list has a limit of 5 elements. So if someone adds 6th element in the URL the list would simply ignore it. So I am trying to use a counter but the logic is not working as desired. I am using While loop to achieve this. So if list size is smaller than 5 set the agencyCds otherwise just return the list.
private List<IUiIntegrationDto> generateViewIntegrationReportData(ESignatureIntegrationConfig eSignConfig) throws Exception {
int counter = 1;
if(eSignConfig.getAdditionalAgencyCds() != null ) {
List<String> combinedAgencyCds = new ArrayList<String>();
for(String agencyCd : eSignConfig.getAgencyCd()) {
combinedAgencyCds.add(agencyCd);
}
StringTokenizer token = new StringTokenizer(eSignConfig.getAdditionalAgencyCds().toString(), StringConstants.COMMA);
while(token.hasMoreTokens()) {
combinedAgencyCds.add(token.nextToken());
}
while(combinedAgencyCds.size() < 5) {
counter = counter + 1;
eSignConfig.setAgencyCd(combinedAgencyCds);
}
// eSignConfig.setAgencyCd(combinedAgencyCds);
}
List<IUiIntegrationDto> intgList = getUiIntegrationManager().retrieveUiIntegrationReportData(eSignConfig.getAgencyCd(), eSignConfig.getCreatedDays(),
eSignConfig.getLob(), eSignConfig.getTransactionStatus(), eSignConfig.getAccounts(), eSignConfig.getSortKey(), eSignConfig.getSortOrder());
return intgList;
}
I am not completely sure about this logic if it is correct or if there is nay better approach.
Thanks
Try this instead of the last while in your code:
if(combinedAgencyCds.size() <= 5) {
eSignConfig.setAgencyCd(combinedAgencyCds);
} else {
eSignConfig.setAgencyCd(combinedAgencyCds.subList(0, 5));
}
The full combined list will then be used if it is less than 5 in size. Otherwise, only the first 5 elements are used.
Edit: Or even better:
eSignConfig.setAgencyCd(combinedAgencyCds.subList(0, Math.min(5, combinedAgencyCds.size())));
Ok so let's break down what your code is currently doing.
int counter = 1;
while(combinedAgencyCds.size() < 5) {
counter = counter + 1;
eSignConfig.setAgencyCd(combinedAgencyCds);
}
This snippet of code has a couple things wrong best I can tell. First, this loop has the possibility of running forever or not at all. Because combinedAgencyCds is never being manipulated, the size won't ever change and the logic being checked in the while loop never does anything. Second, there's a more efficient loop for doing this, assuming you don't need the counter variable outside of its usage in the while loop and that is using for loops.
Example syntax is as follows:
for (int i = 0; i < combinedAgencyCds.size(); i++) {
if (i < 5) {
// Do your logic here.
}
else {
break; // Or handle extra values however you want.
}
}
Notice there is no need for the explicit declaration for a counter variable as "i" counts for you.
Now in your actual logic in the loop, I'm not sure what the setAgencyCd method does, but if it simply sets a list variable in the eSignConfig like it appears to, repeating it over and over isn't going to do anything. From what I can see in your code, you are setting a variable with the same value 5 times. If you need any more explanation just let me know and I will be happy to revise the answer.
Tried changing around the for loop condition several times, still get ArrayIndexOutOfBounds when I pass zero as a parameter. Every other number works fine, I am trying to account for zero by setting it equal to zero automatically, am I doing that part incorrectly? Everything compiles and runs fine except for zero.
private static int iterativeCalculation(int userEntry)
{
int iterativeArray[] = new int[userEntry + 1];
iterativeArray[0] = 0;
iterativeArray[1] = 1;
for (int i = 2; i <= userEntry; i++)
{
iterativeArray[i] = (3 * iterativeArray[i - 1]) - (2 * iterativeArray[i - 2]);
iterativeEfficiencyCounter++;
}
return iterativeArray[userEntry];
}
public static void main(String[] args) {
System.out.println(iterativeCalculation(0));
}
Tried debugging my way through the code, still not understanding what is going wrong. Would appreciate any help! Thanks!
When you pass zero as parameter, userEntry + 1 = 1.
But here:
iterativeArray[1] = 1;
You are trying to set the second element's value. Remember that length of array is one less than its actual size. So removing this line will fix it. Or use userEntry + 2 instead and alter your loop accordingly.
EDIT:
If you really want to fix first and second element, then use this instead:
int iterativeArray[] = new int[userEntry + 2];
iterativeArray[0] = 0;
iterativeArray[1] = 1;
This will create an array of adequate base size.
And remember, length you enter in [...] while creating array has to be one more than the actual length you want. Because actual array starts counting from 0.
In your case, you were setting length as 1 (minimum). That would create an array which can store only one element; that is iterativeArray[0] = //something. Anything above that is OutOfBounds.
You are setting iterativeArray[1] = 1; regardless of whether or not there are actually 2 or more items in the array. That will be out of bounds with one element.
I think you should step through the code in debugger to best understand what the problem is. You'll see exactly where it's got a problem if you single-step through the code. This is a fundamental technique and tool.
I have a Collection of objects I need to iterate over. The collection is of variable size. If the collection has more than 1 object in it, I need to perform special processing on objects 2 .. infinity.
What's the preferred method to do this? For example:
int count = 1;
for (CustomObject co : CustomObjectCollection) {
methodAll(co);
if(count > 1) {
methodSpecial(co);
}
count = count++;
}
What you have will work except for one bug: count = count++ does absolutely nothing. count = count + 1 would work, or count++, but count = count++ is a no-op.
You could also just use a boolean flag if you don't specifically need to track the count.
boolean first = true;
for (CustomObject co : CustomObjectCollection) {
methodAll(co);
if (!first) {
methodSpecial(co);
}
first = false;
}
Which to use
Which to use depends on your specific use case. Assuming that you are not looking to optimise every last op-per-second of performance then go with the option that declares your intent:
count++ is fine if you want to track or use the count outside the loop (as per Louis' answer, it is count++, not count = count++)
for(int i = 0; i < collection.size(); i++) is good as well if the collection supports a get(i) operation. This also lets you skip the first item by initialising int i to a different index. It might be used if you don't want the extra count variable hanging around outside the loop.
the above boolean first = true; (or the inverse boolean notFirst = false;) highlight that you want to treat the first and subsequent elements differently
If you have the List interface on your collection and want to skip a set number of elements, then subList is a good option
Performance
If performance is a concern, then measure it for your platform and implementation, but from general experience, from slowest to fastest, with <=20x difference between the first and last:
for each loop with iterator: slowest
for(int i = 0; i < list.size() ; i++)
declare int size = list.size() then a for(int i = 0; i < size; i++) loop
However these speed results depend on so many things that unless performance is a design goal or an identified issue go with the iterator until you have a reason to use one of the other two - the iterator is generally fast enough.
I think a better way to do this would be to work directly with the iterator like so:
Iterator<CustomObject> it = customObjectCollection.iterator();
if(it.hasNext()) { //first pass don't call methodSpecial
methodAll(it.next());
}
CustomObject customObj;
while(it.hasNext()) { //all the rest 2..infinity
customObj = it.next();
methodAll(customObj);
methodSpecial(customObj);
}
customObj = null; // for garbage collection
This way you don't have to check each iteration if this is the first run or not.
This will work with any Iterable (which is already needed for a foreach loop anyway).
Note: If this is not an ordered collection you might get different elements in the first iteration then you might expect.
Hey guys, recently posted up about a problem with my algorithm.
Finding the numbers from a set which give the minimum amount of waste
Ive amended the code slightly, so it now backtracks to an extent, however the output is still flawed. Ive debugged this considerablychecking all the variable values and cant seem to find out the issue.
Again advice as opposed to an outright solution would be of great help. I think there is only a couple of problems with my code, but i cant work out where.
//from previous post:
Basically a set is passed to this method below, and a length of a bar is also passed in. The solution should output the numbers from the set which give the minimum amount of waste if certain numbers from the set were removed from the bar length. So, bar length 10, set includes 6,1,4, so the solution is 6 and 4, and the wastage is 0. Im having some trouble with the conditions to backtrack though the set. Ive also tried to use a wastage "global" variable to help with the backtracking aspect but to no avail.
SetInt is a manually made set implementation, which can add, remove, check if the set is empty and return the minimum value from the set.
/*
* To change this template, choose Tools | Templates
* and open the template in the editor.
*/
package recursivebacktracking;
/**
*
* #author User
*/
public class RecBack {
int WASTAGE = 10;
int BESTWASTAGE;
int BARLENGTH = 10;
public void work()
{
int[] nums = {6,1,2,5};
//Order Numbers
SetInt ORDERS = new SetInt(nums.length);
SetInt BESTSET = new SetInt(nums.length);
SetInt SOLUTION = new SetInt(nums.length);
//Set Declarration
for (int item : nums)ORDERS.add(item);
//Populate Set
SetInt result = tryCutting(ORDERS, SOLUTION, BARLENGTH, WASTAGE);
result.printNumbers();
}
public SetInt tryCutting(SetInt possibleOrders, SetInt solution, int lengthleft, int waste)
{
for (int i = 0; i < possibleOrders.numberInSet(); i++) // the repeat
{
int a = possibleOrders.min(); //select next candidate
System.out.println(a);
if (a <= lengthleft) //if accecptable
{
solution.add(a); //record candidate
lengthleft -= a;
WASTAGE = lengthleft;
possibleOrders.remove(a); //remove from original set
if (!possibleOrders.isEmpty()) //solution not complete
{
System.out.println("this time");
tryCutting(possibleOrders, solution, lengthleft, waste);//try recursive call
BESTWASTAGE = WASTAGE;
if ( BESTWASTAGE <= WASTAGE )//if not successfull
{
lengthleft += a;
solution.remove(a);
System.out.println("never happens");
}
} //solution not complete
}
} //for loop
return solution;
}
}
Instead of using backtracking, have you considered using a bitmask algorithm instead? I think it would make your algorithm much simpler.
Here's an outline of how you would do this:
Let N be number of elements in your set. So if the set is {6,1,2,5} then N would be 4. Let max_waste be the maximum waste we can eliminate (10 in your example).
int best = 0; // the best result so far
for (int mask = 1; mask <= (1<<N)-1; ++mask) {
// loop over each bit in the mask to see if it's set and add to the sum
int sm = 0;
for (int j = 0; j < N; ++j) {
if ( ((1<<j)&mask) != 0) {
// the bit is set, add this amount to the total
sm += your_set[j];
// possible optimization: if sm is greater than max waste, then break
// out of loop since there's no need to continue
}
}
// if sm <= max_waste, then see if this result produces a better one
// that our current best, and store accordingly
if (sm <= max_waste) {
best = max(max_waste - sm);
}
}
This algorithm is very similar to backtracking and has similar complexity, it just doesn't use recursion.
The bitmask basically is a binary representation where 1 indicates that we use the item in the set, and 0 means we don't. Since we are looping from 1 to (1<<N)-1, we are considering all possible subsets of the given items.
Note that running time of this algorithm increases very quickly as N gets larger, but with N <= around 20 it should be ok. The same limitation applies with backtracking, by the way. If you need faster performance, you'd need to consider another technique like dynamic programming.
For the backtracking, you just need to keep track of which element in the set you are on, and you either try to use the element or not use it. If you use it, you add it to your total, and if not, you proceeed to the next recursive call without increasing your total. Then, you decrement the total (if you incremented it), which is where the backtracking comes in.
It's very similar to the bitmask approach above, and I provided the bitmask solution to help give you a better understanding of how the backtracking algorithm would work.
EDIT
OK, I didn't realize you were required to use recursion.
Hint1
First, I think you can simplify your code considerably by just using a single recursive function and putting the logic in that function. There's no need to build all the sets ahead of time then process them (I'm not totally sure that's what you're doing but it seems that way from your code). You can just build the sets and then keep track of where you are in the set. When you get to the end of the set, see if your result is better.
Hint2
If you still need more hints, try to think of what your backtracking function should be doing. What are the terminating conditions? When we reach the terminating condition, what do we need to record (e.g. did we get a new best result, etc.)?
Hint3
Spoiler Alert
Below is a C++ implementation to give you some ideas, so stop reading here if you want to work on it some more by yourself.
int bestDiff = 999999999;
int N;
vector< int > cur_items;
int cur_tot = 0;
int items[] = {6,1,2,5};
vector< int > best_items;
int max_waste;
void go(int at) {
if (cur_tot > max_waste)
// we've exceeded max_waste, so no need to continue
return;
if (at == N) {
// we're at the end of the input, see if we got a better result and
// if so, record it
if (max_waste - cur_tot < bestDiff) {
bestDiff = max_waste - cur_tot;
best_items = cur_items;
}
return;
}
// use this item
cur_items.push_back(items[at]);
cur_tot += items[at];
go(at+1);
// here's the backtracking part
cur_tot -= items[at];
cur_items.pop_back();
// don't use this item
go(at+1);
}
int main() {
// 4 items in the set, so N is 4
N=4;
// maximum waste we can eliminiate is 10
max_waste = 10;
// call the backtracking algo
go(0);
// output the results
cout<<"bestDiff = "<<bestDiff<<endl;
cout<<"The items are:"<<endl;
for (int i = 0; i < best_items.size(); ++i) {
cout<<best_items[i]<<" ";
}
return 0;
}