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recursive game plan function

I want to write a simple recursive function for a game plan in java.
I have 2^k teams and want an output like this (e.g. 4 teams):
(team)(day1)(day2)(day3)
1 2 3 4
2 1 4 3
3 4 1 2
4 3 2 1
My idea was to call the function recursive with half of its original size, but I can't figure out how to code it properly. If called with n/2, the output has to go into the upper left corner of the plan, the output PLUS n/2 has to go to the lower left corner and the rest is symmetric to the center.
Thanks
My code so far
public void plan(int size) {
if(size==2){}
else{}
}

make a Set for each day (unique)
then use n(max number)
itterate a loop n times and each itteration
newRandomNumber % n (fetch a random number a limit it to 0 to (n-1)
now add the (generatedValue+1) to the set
if it already exists (do check ) then increment the value till its new value then add to set
note: i dont get ur symetric requirement

Java data structure representation

I'm having quite a bit of problem deciding how to represent a certain problem. Basically, I'll need a list/vector with a given number from args[1]. Each of them will have from 0 up till some args[2].
I'll need to make an algorithm to search all possible combinations of arguments from the elements within the first one.
As an example:
1 -> 3 5
2 -> 1 3 4 5
3 -> 2
4 -> 2 5 1
5 -> 1 3 4
From this example, I'll have to generate all the combination of paths, like from 1 you can go to 3 or 5 and so on and cant go back through the same mini path so I'm guessing I'll have to figure out some recursive way to fully search all paths.
My first attemp was to make a vector for 1 to 5 with arraylist for each element but I'm having problem initializing and accessing values withing the lists.
for (int i=0; i < total; i++) {
matrix[i] = new ArrayList<int>(total);
}
for (int i=0; i < total_paths; i++) {
matrix[Integer.parseInt(args[i])].add(Integer.parseInt(args[i]));
}
Any suggestions are welcome.

You can't have an ArrayList<int>; only an ArrayList<Integer>. However, Java will automatically convert between int and Integer for you in most cases.

You can't have an ArrayList<int> - it has to be ArrayList<Integer>.
Lists can only store Objects, such as Integer, String, etc. In Java, int is a primitive type, not a true Object.
Your second code block will boil down to:
matrix[1].add(1);
(for example - the same value as an array index and a List entry) - is that what you meant? It's not entirely clear how you are passing the values in...

Reversing a part of an array (or any other data structure )

I want to reverse an array (or any other data structure) but because this operation is going to be done on the array for n times , im looking for the best solution possible, I have the sorted array , which is gotten in O(nlgn) time , i start looking for first element in the sorted array , inside the unsorted array ( which is equal to finding the smallest key in the unsorted array ) then I reverse the array from the beginning to the index of that value , then i do the same for the rest , find the second smallest value's index , and reverse the array again , from the second index to the end of the array and so on :
for example , consider this array :
*‫2 6 (*1*) 5 4 *3‬ // array is reversed from its 0th index to the 3rd index (0 based)
1 *5* 4 3 6 (*2*) // array is reversed from its 1st index (0 based ) to the 5th index
1 2 *6* *3* 4 5 // and ...
1 2 3 *6* *4* 5
1 2 3 4 *6* *5*
1 2 3 4 5 6
well i have to sort the array in order to have the values im looking for in the unsorted array , it'll take o(nlgn) time , and doing the algorithm above , will take o(n^2) ,any idea to make it more quick , to be done in o(nlgn) time ? so the question is reversing a sub array of the array in the least time Order , cause it's done for many times in large sized arrays. ( I can get the indices i and j ( which are the first and last index of the sub array ) in O (n) time cause i have the sorted array and i'll just look up the numbers in the unsorted array ) so im looking for the best time order for reversing an array from it's ith index to it's jth index .
thanks in advance

Here comes an O(n) solution (i think, reading your description was hard). It's a data structure wich allows 1) reversing a sub-array in O(1), 2) Getting a value from the array in O(r), where r is the number of reversings that is done, 3) find the index of an element in O(n), where n is the length of the list.
Just store our array as usual, and have a list of ReversedRange(imin, imax) elements. Reversing part of the array is as easy as inserting another element in this list.
Whenever you need to get a value from the modified array at index i, you look through all the ReversedRange for which imin <= i <= imax, and calculate the index j which corresponds to the original array index. You need to check r reversings, so it is O(r).
To get the index i of a value v, look through the original array and find the index j. Done in O(n) time. Do the same traversing of the ReversedRanges, only in the oppsite direction to calculate i. Done in O(r) time, total O(n+r) which is O(n).
Example
Consider the following list: 0 1 2 3 4 5 6 7 8 9. Now say we reverse the list form indexes 1 through 6, and then from 0 through 5. So we have:
I 0 1 2 3 4 5 6 7 8 9
| |
II 0 6 5 4 3 2 1 7 8 9
| |
III 2 3 4 5 6 0 1 7 8 9
No let us map the index i = 2 to the original array. From the graph we see that we should end up with III(2) = 4
1) since i = 2 is in [0, 5] we calculate
i <- 5 - (i - 0) = 5 - 2 = 3
2) since i = 3 is in [1, 6] we calculate
i <- 6 - (i - 1) = 6 - 2 = 4
3) there are no more ranges, we are done with result 4!

(Possibly) memory heavy...
Why don't you maintain two Collections. Traverse the original collection until you have found your reverse point. Then traverse backwards from there with an iterator or indexing if you have it (ArrayList) adding each element to a new collection. Then merge the two collections (The reversed portion and the previous untouched portion). Repeat this until finished.

Seems like a very simple answer to a complicated question so maybe I'm missing something. If you are looking for an efficient way to reverse part of an array then something like the following should work. You could easily make it generic of course.
// endIndex and startIndex are inclusive here
int half = startIndex + ((endIndex + 1) - startIndex) / 2;
int endCount = endIndex;
for (int startCount = startIndex; startCount < half; startCount++) {
int store = array[startCount];
array[startCount] = array[endCount];
array[endCount] = store;
endCount--;
}
Seems like the rest of your code would be much more complex than this. Not sure what the O() is here because it is not doing comparisons which are traditionally the measure but this is doing 1.5 x N assignments to reverse the array. I don't see any way that this can be done faster.

How does one output a new array, containing all numbers of a previous array that occur at most k times

How does one create a new array, containing all numbers of the array that occur at most k times in java?
For example, if the array was:
{1,4,4,3,4,3,5,2,5,1,5}
and k = 2, the new array would be:
{1,3,2}.
I assume that I would need to compare the elements within the arrays and add store the number of times it occurs as a variable. Then I would compare that variable to k and if it is smaller or equal to it, it will add it to a new arraylist. I can't really implement this tho.
Thanks

There are many ways to do this, depending on what your constraints are.
If you don't need to worry about memory constraints, you can solve this problem quite easily by using a HashMap that maps from array elements to their frequencies. The algorithm would work by scanning across the array. For each element, if that element is already in the HashMap, you update the key/value pair in the HashMap to increment the frequency of that element. Otherwise, if the element hasn't been seen, you update the frequency to be 1. Once you've finished populating the HashMap, you can then iterate across the map and copy over all elements that have frequency at most k into a new ArrayList. For example:
for(Map.Entry<T, Integer> entry: myMap.entrySet()) {
if (entry.getValue() <= k)
myArrayList.add(entry.getKey());
}
This runs in O(n) time and O(n) memory, which is quite good.
If you don't want to store everything in memory, another option would be to sort the array in O(n log n) time and O(log n) space using Arrays.sort. Once you've sorted the array, all of the copies of each value will be stored consecutively and you can more easily count their frequency. For example, after sorting the array you mentioned in your original post, you would get the array
1 1 2 3 3 4 4 4 5 5 5
From here, it should be much easier to determine how many copies of each element there are. You could do this by walking across the array, keeping track of the element you're currently looking for and how many copies there are. Whenever you see a new element, if it matches the current element, you increment the counter. If not, you reset the element you're checking to be the new element and set the counter back to one. For example, with the above array, you might start off by looking at this element:
1 1 2 3 3 4 4 4 5 5 5
^
Element: 1
Frequency: 1
Now, you look at the next element of the array. Because it's a 1, you increment the frequency count to 2:
1 1 2 3 3 4 4 4 5 5 5
^
Element: 1
Frequency: 2
When you look at the next element, you'll see that it's a 2. This means that there aren't any more 1's left. Because there are two copies of 1, you could then append 1 to the resulting array. You'd then reset the counter to 1, leaving this state:
1 1 2 3 3 4 4 4 5 5 5
^
Element: 2
Frequency: 1
The one thing to watch out for when doing this is to remember to handle the case where you visit the final array element. It's important that when you hit the end of the array, if you have at most k copies of the last element, you add the last number to the output array.
Whenever you do this, if you ever find an element that appears at most k times, you can add it to the new array. This second step runs in O(n) time and O(m) space (where m is the total number of elements in the resulting array), so the total complexity is O(n log n) time and O(m + log n) space.
Hope this helps!

You have at least two ways to do it:
Scan linearly the array, updating an hash table which stores the frequency of each encountered number (Average time: O(n), Space: O(n));
Sort the array and keeping a count of how many times you've seen the current element (which will be set to one every time the current number is different from the previous one). Time: O(n lg n), Space: O(m), where m is the number of returned element (assuming to use a O(1) space-complexity sorter, like heapsort).
The first one is time-optimal, the second one is space-optimal.

Sorting an array while moving duplicates to the end?

This was a question in one my friend's programming class.
Q. How do you sort an array of ints and then arrange them such that all duplicate elements appear at the end of the array?
For example, given the input
{5, 2, 7, 6, 1, 1, 5, 6, 2}
The output would be
{1, 2, 5, 6, 7, 1, 2, 5, 6}
Note that the numbers are sorted and duplicate numbers are after 7, which is the maximum in the array.
This has to be achieved with out using any Java library packages/utils.
I suggested to sort the array first using insertion or bubble sort, and then go over the array, perform something like the following :
for (int i = 0; i < nums.length - 2; i++) {
for (int j = i + 1; j < nums.length; j++) {
//current and next are same, move elements up
//and place the next number at the end.
if (nums[i] == nums[j]) {
int temp = nums[j];
for (int k = j; k < nums.length - 1; k++) {
nums[k] = nums[k + 1];
}
nums[nums.length - 1] = temp;
break;
}
}
}
I tried this myself later (and that is how the code above) - As I try this out, I think this could be achieved by using less code, be more efficiently. And may be I gave a wrong advice.
Any thoughts?

Depending on the parameters of your problem, there are many approaches to solving this.
If you are not allowed to use O(n) external memory, then one option would be to use a standard sorting algorithm to sort the array in-place in O(n log n) time, then to run a second pass over it to move the duplicates to the end (as you've suggested). The code you posted above takes O(n2) time, but I think that this step can be done in O(n log n) time using a slightly more complicated algorithm. The idea works in two steps. In the first step, in O(n log n) time you bring all non-duplicated elements to the front in sorted order and bring all the duplicates to the back in non-sorted order. Once you've done that, you then sort the back half of the array in O(n log n) time using the sorting algorithm from the first step.
I'm not going to go into the code to sort the array. I really love sorting, but there are so many other good resources on how to sort arrays in-place that it's not a good use of my time/space here to go into them. If it helps, here's links to Java implementations of heapsort, quicksort, and smoothsort, all of which runs in O(n log n) time. Heapsort and smoothsort use only O(1) external memory, while quicksort can use O(n) in the worst case (though good implementations can limit this to O(log n) using cute tricks).
The interesting code is the logic to bring all the non-duplicated elements to the front of the range. Intuitively, the code works by storing two pointers - a read pointer and a write pointer. The read pointer points to the next element to read, while the write pointer points to the location where the next unique element should be placed. For example, given this array:
1 1 1 1 2 2 3 4 5 5
We start with the read and write pointers initially pointing at 1:
write v
1 1 1 1 2 2 3 4 5 5
read ^
Next, we skip the read pointer ahead to the next element that isn't 1. This finds 2:
write v
1 1 1 1 2 2 3 4 5 5
read ^
Then, we bump the write pointer to the next location:
write v
1 1 1 1 2 2 3 4 5 5
read ^
Now, we swap the 2 into the spot held by the write pointer:
write v
1 2 1 1 1 2 3 4 5 5
read ^
advance the read pointer to the next value that isn't 2:
write v
1 2 1 1 1 2 3 4 5 5
read ^
then advance the write pointer:
write v
1 2 1 1 1 2 3 4 5 5
read ^
Again, we exchange the values pointed at by 'read' and 'write' and move the write pointer forward, then move the read pointer to the next unique value:
write v
1 2 3 1 1 2 1 4 5 5
read ^
Once more yields
write v
1 2 3 4 1 2 1 1 5 5
read ^
and the final iteration gives
write v
1 2 3 4 5 2 1 1 1 5
read ^
If we now sort from the write pointer to the read pointer, we get
write v
1 2 3 4 5 1 1 1 2 5
read ^
and bingo! We've got the answer we're looking for.
In (untested, sorry...) Java code, this fixup step might look like this:
int read = 0;
int write = 0;
while (read < array.length) {
/* Swap the values pointed at by read and write. */
int temp = array[write];
array[write] = array[read];
array[read] = temp;
/* Advance the read pointer forward to the next unique value. Since we
* moved the unique value to the write location, we compare values
* against array[write] instead of array[read].
*/
while (read < array.length && array[write] == array[read])
++ read;
/* Advance the write pointer. */
++ write;
}
This algorithm runs in O(n) time, which leads to an overall O(n log n) algorithm for the problem. Since the reordering step uses O(1) memory, the overall memory usage would be either O(1) (for something like smoothsort or heapsort) or O(log n) (for something like quicksort).
EDIT: After talking this over with a friend, I think that there is a much more elegant solution to the problem based on a modification of quicksort. Typically, when you run quicksort, you end up partitioning the array into three regions:
+----------------+----------------+----------------+
| values < pivot | values = pivot | values > pivot |
+----------------+----------------+----------------+
The recursion then sorts the first and last regions to put them into sorted order. However, we can modify this for our version of the problem. We'll need as a primitive the rotation algorithm, which takes two adjacent blocks of values in an array and exchanges them in O(n) time. It does not change the relative order of the elements in those blocks. For example, we could use rotation to convert the array
1 2 3 4 5 6 7 8
into
3 4 5 6 7 8 1 2
and can do so in O(n) time.
The modified version of quicksort would work by using the Bentley-McIlroy three-way partition algortihm (described here) to, using O(1) extra space, rearrange the array elements into the configuration shown above. Next, we apply a rotation to reorder the elements so that they look like this:
+----------------+----------------+----------------+
| values < pivot | values > pivot | values = pivot |
+----------------+----------------+----------------+
Next, we perform a swap so that we move exactly one copy of the pivot element into the set of elements at least as large as the pivot. This may have extra copies of the pivot behind. We then recursively apply the sorting algorithm to the < and > ranges. When we do this, the resulting array will look like this:
+---------+-------------+---------+-------------+---------+
| < pivot | dup < pivot | > pivot | dup > pivot | = pivot |
+---------+-------------+---------+-------------+---------+
We then apply two rotations to the range to put it into the final order. First, rotate the duplicate values less than the pivot with the values greater than the pivot. This gives
+---------+---------+-------------+-------------+---------+
| < pivot | > pivot | dup < pivot | dup > pivot | = pivot |
+---------+---------+-------------+-------------+---------+
At this point, this first range is the unique elements in ascending order:
+---------------------+-------------+-------------+---------+
| sorted unique elems | dup < pivot | dup > pivot | = pivot |
+---------------------+-------------+-------------+---------+
Finally, do one last rotation of the duplicate elements greater than the pivot and the elements equal to the pivot to yield this:
+---------------------+-------------+---------+-------------+
| sorted unique elems | dup < pivot | = pivot | dup > pivot |
+---------------------+-------------+---------+-------------+
Notice that these last three blocks are just the sorted duplicate values:
+---------------------+-------------------------------------+
| sorted unique elems | sorted duplicate elements |
+---------------------+-------------------------------------+
and voila! We've got everything in the order we want. Using the same analysis that you'd do for normal quicksort, plus the fact that we're only doing O(n) work at each level (three extra rotations), this works out to O(n log n) in the best case with O(log n) memory usage. It's still O(n2) in the worst case with O(log n) memory, but that happens with extremely low probability.
If you are allowed to use O(n) memory, one option would be to build a balanced binary search tree out of all of the elements that stores key/value pairs, where each key is an element of the array and the value is the number of times it appears. You could then sort the array in your format as follows:
For each element in the array:
If that element already exists in the BST, increment its count.
Otherwise, add a new node to the BST with that element having count 1.
Do an inorder walk of the BST. When encountering a node, output its key.
Do a second inorder walk of the BST. When encountering a node, if it has count greater than one, output n - 1 copies of that node, where n is the number of times it appears.
The runtime of this algorithm is O(n log n), but it would be pretty tricky to code up a BST from scratch. It also requires external space, which I'm not sure you're allowed to do.
However, if you are allowed external space and the arrays you are sorting are small and contain small integers, you could modify the above approach by using a modified counting sort. Just replace the BST with an array large enough for each integer in the original array to be a key. This reduces the runtime to O(n + k), with memory usage O(k), where k is the largest element in the array.
Hope this helps!

a modified merge sort could do the trick: on the last merge pass keep track of the last number you pushed on the front of result array and if the lowest of the next numbers is equal add to the end instead of front

Welcome to the world of Data Structures and Algorithms. You're absolutely right in that you could sort that faster. You could also do it a dozen different ways. PHD's are spent on this stuff :)
Here's a link where you can see an optimized bubble sort
You might also want to check out Big O Notation
Have fun and good luck!

Use quicksort to sort the array. When implementing the sort you can modify it slightly by adding all duplicates to a seperate duplicate array. When done simply append the duplicate array to the end of the sorted array.