I have to implement a Selection Sort in Java according to these parameters:
Implement a variation to the SelectionSort that locates both the smallest and largest elements while scanning the list and positions them at the beginning and the end of the list, respectively. On pass number one, elements x0,...,xn-1 are scanned; on pass number two, elements x1,...,xn-2 are scanned; and so on.
I am passing the method an array of size 32, and when I print the array it is not sorted. What's the matter with my code?
static void selectionSort() {
scramble();
int smallIndex = 0; //index of smallest to test
int largeIndex = array.length - 1; //index of largest to test
int small = 0; //smallest
int large; //largest
int smallLimit = 0; //starts from here
int largeLimit = array.length - 1; //ends here
int store; //temp stored here
int store2;
for(int i = 0; i < array.length/2; i++) { //TODO not working...
small = array[smallLimit];
large = array[largeLimit];
for(int j = smallLimit; j <= largeLimit; j++) {
if(array[j] < small) {
smallIndex = j;
small = array[j];
}
else if(array[j] > large) {
largeIndex = j;
large = array[j];
}
}
store = array[smallLimit];
store2 = array[smallIndex];
array[smallLimit] = store2;
array[smallIndex] = store;
store = array[largeLimit];
array[largeLimit] = array[largeIndex];
array[largeIndex] = store;
smallLimit++;
largeLimit--;
}
print();
}
Think about the extreme cases: what happens when the largest or smallest item is found at smallLimit or largeLimit. When that happens you have two problems:
largeIndex and smallIndex are not set. They maintain their values from a previous iteration.
Swapping the smallest item to its correct place moves the largest item. The second swap moves the smallest item where the largest should go, and the largest ends up in a random location. Step through the code on paper or in a debugger if you find this hard to visualize.
These problems are easy to fix. You could have avoided the problem following a few guidelines:
Use fewer moving parts. You can always get the value of small using smallIndex, if you just used smallIndex there would be no danger of different variables falling out of step.
Declare the variables in the smallest possible scope. If smallIndex was declared in the loop body and not outside the compiler would have told you there's a chance it was not set before the swap.
Destructive updates like the first swap here can always make a previous calculation obsolete. When you can't avoid this from happening, look for ways two updates can step on each other's toes.
Like #Joni, clearly pointed out, there is big caveat with swapping two elements twice during a traversal of the array. Since you have to implement the sorting algorithm in-place, you need to take into account the positions of the elements to be swapped as it happens in succession.
Another limiting case that you need to see is when there are just three elements left i.e. the last iteration of the for loop. This is how I would go about it:
store = array[smallLimit];
store2 = array[smallIndex];
array[smallLimit] = small;
array[smallIndex] = store;
smallLimit++;
//No problem with swapping the first two elements
store = array[largeLimit];
//However the first swap might cause the other elements to shift
//So we do this check
if((array[largeIndex] == large))
{array[largeLimit] = array[largeIndex];
array[largeIndex] = store;
largeLimit--;
}
//Just a limiting case, where amongst the last three elements, first swap happens.
//The smallest element is in place, just take care of the other two elements.
if(largeLimit - smallLimit == 1){
if(array[largeLimit] != large){
array[smallLimit] = array[largeLimit];
array[largeLimit] = large;
largeLimit--;
}
}
Working DEMO for the snippet mentioned above, building upon your code. Hope it gets you started in the right direction.
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);
for(int i = 0; i < bag.length; i++)
{
if(bag[i].equals(a))
{
tmp = bag[i];
bag[i] = bag[bag.length-1];
bag[bag.length-1] = tmp;
numElements--;
break;
}
}
The goal of this is to find an object in the array and then remove it? is it possible??
Changing the length of an array is not possible. Recall that array is a static data structure whose size is determined before hand. Increasing or decreasing is not supported in this data structure. The fact that one has to increase or decrease the size depending on the usecase means that they have picked up the wrong data structure. They should perhaps go with an ArrayList.
Anyway, coming back to your question, you can simulate the 'size decrease' by maintaining a variable which you let track the array index and decrease the size of this variable. This lets you give the impression of shrinking the array.
The code you have provided does the same. Note however, that you should be using this modified index to track the contents of your array.
for(int i = 0; i < bag.length; i++)
{
if(bag[i].equals(a))
{
tmp = bag[i];
bag[i] = bag[bag.length-1];
bag[bag.length-1] = tmp;
numElements--;
break;
}
}
Whenever a particular bag at a given index equals to the item under question i.e., 'a', we swap elements so that the current bag element to be removed moves to the last and also we reduce the size of our new index - numElements by 1 to simulate this.
If you have the full code with you, please consider adding the following snippet at the end of that program to understand this more:
// Simulation of the array shrinking.
for(int i = 0; i < numElements; i++)
{
System.out.println( bag[i] );
}
// Movement of uninteresting elements to the end of the array.
for(int i = 0; i < bag.length; i++)
{
System.out.println( bag[i] );
}
It's not possible to change the length of an array. You can overwrite the element you wish to remove with the last element of the array and then copy the first bag.length - 1 elements of your array to a new array whose length is bag.length - 1.
for(int i = 0; i < bag.length; i++) {
if(bag[i].equals(a)) {
bag[i] = bag[bag.length-1];
bag = Arrays.copyOf (bag, bag.length - 1);
break;
}
}
public static String[] removeElements(String[] input) {
List<String> result = new ArrayList<String>();
String deleteValue = "somevalue";
for(String item : input)
if(!deleteValue .equals(item))
result.add(item);
return result.toArray(input);
}
This is one method you can fit this into your program.
You cannot decrease the size of an array. okay no problem! you can create your own data structure which supports that right?
Now, create a class named say MyArray with functions like increaseLenght(int) and decreseLength(int). Try it if you want to, will be fun for sure..
You cannot reduce the size of an array. Arrays are fixed length. What you can do is have a variable that indicates how many entries of the array you are using. This is what you are doing with numElements. The standard class ArrayList is implemented like this. The data is kept in an array and a private field size is used. With an ArrayList, when you remove an element, all the elements to the right are shifted left. However I also like your idea.
I would suggest 2 changes.
Make the last element null instead. If you are removing the element, why does it still need to be in the array?
Use numElements - 1 rather than bag.length-1 as the array could be bigger.
With these changes it becomes:
for(int i = 0; i < bag.length; i++)
{
if(bag[i].equals(a))
{
bag[i] = bag[numElements-1];
bag[numElements-1] = null;
numElements--;
break;
}
}
I have an ArrayList of String, and I would like to retrieve the first and last result of the names after calculating the order of alphabets. Below is my code snippet:
ArrayList<String> list = new ArrayList<String>(20);
list.add("Charles Darwin");
list.add("Albert Einstein");
list.add("Issac Newton");
list.add("Tony Hoare");
list.add("Grace Hopper");
list.add("Edgar Dijkstra");
list.add("Ada Lovelace");
list.add("Charles Babbage");
list.add("Stephen Hawking");
String biggest = "";
String smallest = "";
for (int i = 0; i < list.size(); i++) {
String first = list.get(i);
for (int j = 0; j < list.size(); j++) {
String second = list.get(j);
if (!first.equalsIgnoreCase(second)) {
if (first.compareToIgnoreCase(second)>0){
biggest=first;
}
if (first.compareToIgnoreCase(second)<0){
smallest=first;
}
}
}
}
System.out.println(biggest);
System.out.println(smallest);
I am able to retrieve every value for comparison, however, the results are always showing Stephen Hawking as the biggest and smallest.
My desired results are Ada Lovelace as biggest and Tony Hoare as smallest.
You can use Collections.min / max
Your conditional statements seem to wrong.
if (first.compareToIgnoreCase(second)>0){
biggest=first;
}
if (first.compareToIgnoreCase(second)<0){
smallest=first;
}
You're comparing the element in the outer loop to the element in the inner loop. You never make a comparison against the biggest and smallest.
This should help you find the biggest and smallest String in your list.
String biggest = list.get(0);
String smallest = list.get(0);
for (int i = 1; i < list.size(); i++) {
if(list.get(i).compareToIgnoreCase(biggest) > 0)
biggest = list.get(i);
if(list.get(i).compareToIgnoreCase(smallest) < 0)
smallest = list.get(i);
}
Alternatively, you can use Collections.min() and max() as stated in one of the other answers.
are your required to use List ? You might want to see http://docs.oracle.com/javase/7/docs/api/java/util/SortedSet.html. then you can use first() and last() method
I'm new to Java, but I can still recognize multiple issues with this code:
Why are you hard coding initial capacity to 20?
Why are you using indexed loops instead of for each?
Why are you using a nested loop to find min/max?
Use else instead of running the same comparison twice
If first always equals second biggest and smallest will remain uninitiated
And last, and directly dressing your question, all your code does is finding if the last item is greater & smaller than any of the other items in the list, not all of them, since you keep ignoring previous findings and not using biggest/smallest as a condition in any of your comparisons.
This is my minHeap algorithm however it doesn't function as expected:
public static int [] fixheap(int heap[], int n, int i){
int j=2*i;
int weight = heap[i];
while (j<=n){
if((j<n) && heap[j] > heap[j+1])
j++;
if(weight <= heap[j]) break;
else
heap[j/2] = heap[j];
j=j*2;
}
heap[j/2]= weight;
return heap;
}
public static void makeheap(int heap[], int n){
for (int i=n/2; i>=0; i--){
fixheap(heap, n ,i);
}
}
When the data elements are added in various orders the algorithm returns incorrect minHeaps. Can anyone see any apparent problems for this minimum heap algorithm?
You are comparing the wrong elements of the array for forming the heap. Try to dry run your program
As the array starts from the index 0, you should take i=n/2-1 initially here.
public static void makeheap(int heap[], int n){
for (int i=n/2 - 1; i>=0; i--){
fixheap(heap, n ,i);
}
}
And then you will have to change your fixheap function to get the correct value for j
j = i*2 + 1
I believe that the way you find the parent and/or the children is incorrect.
Think about it, if the left children is at index1 and the right at index2, how do I get to their parents at index0?
What about finding index0's children (index1 and index2)?
The folllwing code is in python, but I will highlight the lines that do the heavy lifting, and in the process hopefully present a different solution of creating a min-heap
heapArray = []
for heapKey in someArray:
insert(heapArray, int(heapKey))
return heapArray;
def insert(heapArray, heapKey):
heapArray.append(heapKey)
index_of_key = len(heapArray)-1
parent_index_of_key = (index_of_heap_key-1)/2
while index_of_key>0:
if(heapKey < heapArray[parent_index_of_key]):
__swap(index_of_key, parent_index_of_key, heapArray)
index_of_key = parent_index_of_key;
parent_index_of_key = (index_of_key-1)/2
else:
break # we have reached the right spot
In the above example, we re-create the heap (yes that means more memory, but for illustration purposes this maybe a good start). As we create the heap, we simply check the value of the newly inserted key with its parent (via parent_index_of_key).
If the parent is larger than its child, then we swap the value and update the indexes (both for the swapped key and its new parent). We repeat that process until we have either reached the top of the heap or if the heap key can't go any further up the chain
The in-place swaps are clearly more efficient, but the above is more intuitive and easy to follow. Clearly, I won't use the above code, where memory is a large constraint, but would consider it for cases where code conciseness and clarity trump memory utilization.
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;
}