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Divide Array into 2 equal sum parts using only one loop, need alternative approach

Usually what we do is
1.we calculate sum of whole array from left and call it Leftsum
2.then we starts traversing array from Right end and start adding
every element into RIGHTSUM, and subracting every element from
LEFTSUM
3. untill LEFTSUM=RIGHTSUM,then we assign current value of iterator
variable I to splitpoint
4.Now array before SPLITPOINT index and from SPLITPOINT index to end have equal sum
MAIN PROBLEM------>
but this will work if array is ((3 , 5 , 10) , (3, 10 , 5)) but not on (3 , 5 , 5 , 3, 10 , 10) because we cannot find SPLITPOINT here whose left and right array sum is equal
answer of second part should also be (3,5,10) (5,3,10)
Again sum of 2 parts should be equal not length
this is an easy illustation example above

Suppose the total for the whole sequence is N. You're effectively looking for a sub-sequence that sums to N/2. (Note: if N is odd, there is no such sub-sequence!)
This is the "subset sum" problem. There's a decent Java solution for that at Finding all possible combinations of numbers to reach a given sum.

2D Array containing Arrays

I'm working on a discrete math problem in Java where I'm trying to find r-permutations up to n with repetition. This means if r = 4, n = 2 we can have:
1 1 1 1
1 1 1 2
1 1 2 1
1 1 2 2, etc.
Repetition means 1112 is a unique combination, however 1211 is not because 1112 already contains 3 1's and 1 2.
Therefore I wrote code to iterate through all the permutations, and every iteration is stored in an array. There is a "Master Array" which I planned on being a 2D Array where the first dimension are all unique arrays or permutations. Therefore all I have to do is create a temp array of the new iteration and use Arrays.sort and compare the temp array to all the indexes in the master array (this works because 1211, 1112, 2111, and 1121 all sort to 1112 in ascending order).
Now, the problem I'm having is storing these arrays into the 2D array Masterkey.
I've tried:
Masterkey[count] = array;, but when I print(Masterkey[count][0]) only a single integer is being output rather than the array. (array is the newest permutation)
Masterkey[count] = Arrays.copyOf(array, array.length);, once again prints a single integer rather than the array.
Masterkey[count] = array.clone();, also prints out just an integer.
I may be overlooking something simple, but I'm just struggling to get array to be stored in Masterkey.

Masterkey[count][0] will always print a single integer because Masterkey[count] = array. so Masterkey[count][0] is the same as array[0]. However, if you assign array to Masterkey[count][0] it will print the array.

Create 2 arrays - 1st has random integers, 2nd has unique random integers

I am working on a school homework problem. I need to create 2 int[] arrays. The first array int[10] is filled with random integers. The second array has the same numbers as in the first array, but without any duplicates.
For example, assume my first array is 1,2,2,3,1,5,5,7,9,9. My second array would then be 1,2,3,5,7,9.
Could someone please point me in the right direction to solving this problem.

Put the numbers into a Set. Then retrieve numbers from the Set. Simple! Duplicates will automatically be removed!

I would do the following (assuming that it is homework and you shouldn't be doing anything too complicated)...
Sort the array using java.util.Arrays.sort(myArray); - this will order the numbers, and make sure that all repeating numbers are next to each other.
Loop through the array and keep a count of the number of unique numbers (ie compare the current number to the next number - if they're different, increment the counter by 1)
Create your second int[] array to the correct size (from point 2)
Repeat the same process as point 2, but fill your new array with the unique numbers, rather than incrementing a counter.
This should be enough to get you moving in the right direction. When you have some code, if you still have questions, come back to us and ask.

I recommend using a Set , but here's a way to do it without using a Set. (Note: This will work, but don't ask me about the efficiency of this!)
Have a function like this -
public static boolean isNumberInArray(int[] array, int number)
{
for(int i=0; i<array.length; i++)
{
if(number == array[i])
return true;
}
return false;
}
Now use this function before you make an insert into the new array. I leave you to figure out that part. It's homework after all!

Hints(WATTO explains it better):
a = sorted first array
lastItem = a[0]
append lastItem into new array
for i in 1 to length(a):
if a[i] != lastItem:
append a[i] into new array
lastItem = a[i]

#WATTO Studios has a good approach. Sorting is always useful when duplicates are involved.
I will suggest an alternative method using hash tables:
Create a hashing structure with an integer as key (the number in the original array) and a counter as a value.
Go through the original array and for each number encountered increment it's corresponding counter value in the hash table.
Go through the original array again. For each number check back the hash table. If the counter associated is greater than 1, remove the value and decrement the counter.
Let's see a practical case:
4 5 6 4 1 1 3
First pass will create the following table:
1 -> 2
3 -> 1
4 -> 2
5 -> 1
6 -> 1
Second pass step by step:
4 5 6 4 1 1 3
^
4 has a counter of 2 -> remove and decrement:
1 -> 2
3 -> 1
4 -> 1
5 -> 1
6 -> 1
5 6 4 1 1 3
^
5 has a counter of 1 -> ignore
6 has a counter of 1 -> ignore
4 has a counter of 1 -> ignore
1 has a counter of 2 -> remove and decrement
1 -> 1
3 -> 1
4 -> 1
5 -> 1
6 -> 1
5 6 4 1 3
^
1 has a counter of 1 -> ignore
3 has a counter of 1 -> ignore
Final array:
5 6 4 1 3
There are, of course, more efficient ways to handle the removal (since using an array implies shifting), like inserting the items into a linked list for example. I'll let you decide that. :)
Edit: An even faster approach, requiring a single pass:
Use the same hashing structure as above.
Go through the original array. For each item check the table. If the associated counter is 0, increment it to 1. If it's already 1, remove the item.

How do I use the median of an array as the pivot for quicksort

I have to write a quicksort algorithm that uses the median of the array as the pivot. From my general understanding from what I read in a book, I have to use a select algorithm in which I split the array into n/5 sub arrays, sort each of the sub arrays using insertion sort, find the median, then recursively call the select algorithm to find the median of the medians. I'm not even sure how to start this and I'm pretty confused. the selectMedian algorithm call is supposed to look something like this: SelectMedian(int first, int last, int i) where i is the ith index I want to select (in this case it would be the middle index, so array.length/2). The book I'm reading gives this description of it:
The algorithm in words (if n>1):
1. Divide n elements into groups of 5
2. Find median of each group (use insertion sort for this)
3. Use Select() recursively to find median x of the n/5
medians
4. Partition the n elements around x. Let k = rank(x)
5. if (i == k) then return x
if (i < k) then use Select() recursively to find i-th
smallest element in first partition else (i > k) use
Select() recursively to find (i-k)th smallest element in
last partition.
can anyone assist me in writing this algorithm? thanks!

Would that really be necessary? Why not use the median of three where you select the pivot based on the median of three values, ie. the first, middle and last values.
Or you could even use a random pivot, which will drastically lower the chances of ending up with QuickSort's worst case time of O(N²), which may also be appropriate for your implementation.

I assume you can figure out created n/5 sub-arrays of 5 elements each.
Finding the median of a subarray is fairly easy: you look at each element and find the element which has two smaller elements.
For example, you have 1 4 2 3 5. 1 has no smaller elements. 4 has three smaller elements. 2 has one smaller element. 3 has two smaller elements; this is the one you want.
Now you have found n/5 medians. You want to find the median of the medians, so you run the algorithm again.
Example:
1 7 2 4 9 0 3 8 5 6 1 4 7 2 3
[1 7 2 4 9][0 3 8 5 6][1 4 7 2 3]
findMedian([1 7 2 4 9]) = 4;
findMedian([0 3 8 5 6]) = 5;
findMedian([1 4 7 2 3]) = 3;
[4 5 3]
findMedian([4 5 3]) = 4;
4 is your pivot.
The reason you do this is to try and split your array evenly; if your array is split lopsided, you'll get O(N^2) performance; if your array is split evenly, you get O(NlogN) performance.
Selecting a random pivot means you could get either one - in practice it would balance out to O(NlogN) but a lot of applications want consistent performance, and random quicksort is not consistent from run to run.
The reason we use 5 (instead of 3 or 7) is because we're adding another term of time complexity searching for the median - this term has to be less than O(NlogN) but you want it to be as small as possible. Using 3 gets you O(N^2), using 5 gets you O(NlogN), and 5 is the smallest number for which this is true.
(the algorithm to find the median in linear time was given by Blum, Floyd, Pratt, Rivest and Tarjan in their 1973 paper "Time bounds for selection", and answered a famous open problem)

QUICKSORT2(A, p, r)
if p<r
median=floor((p + r) /2) - p + 1
q=SELECT(A, p, r, median)
q=PARTITION2(A, p, r, q)
QUICKSORT2(A, p, q-1)
QUICKSORT2(A, q+1, r)
PARTITION2(A, p, r, q)
switch between A[r] and A[q]
return PARTITION(A, p, r)

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.