Before Java 8 versions, we can initialize a two-dimensional array using for loop like below. How can I do the below in JDK 1.8 using lambda expressions?
int[][] board = new int[3][3];
int v = 1;
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < 3; j++) {
board[i][j] = v;
v++;
System.out.println(v);
}
}
System.out.println(Arrays.deepToString(board));
Basically, I am looking for an array like below
[[1,6],[2,7],[3,8],[4,9]]
I highly recommend you stick with using the for-loop for initialization of arrays with primitive values. My recommendation is stronger in the case of a multidimensional array.
However, here is the way:
int a = 4; // Number of outer array elements
int b = 2; // Number of inner array elements
int[][] board = IntStream
.range(0, a) // iterate 0..3
.mapToObj(i -> IntStream.range(0, b) // for each iteratoe 0..1
.map(j -> 1 + (i + (a + 1) * j)) // calculate the value
.toArray()) // compose the inner array
.toArray(int[][]::new); // compose the outer array
Note that the IntStream is able to create an array of primitives since it is a sequence of primitive int-valued elements. its method IntStream::toArray reutrns int[].
The composition of the outer array is a bit tricky since int[] is no longer a primitive value but an array itself. There is needed to use a method IntStream::mapToObj which maps int to an object - then the Stream<int[]> is returned and the method Stream::toArray(IntFunction<A[]> generator) converting to array with parameter has to be used since you cannot convert Object[] to int[][].
The parameter passed is simple. int[][]::new is nothing different than i -> new int[a][b].
Use IntStream. This will create a continuous stream of integers from 0 inclusive to n exclusive. So for a 3 x 3 matrix, the IntStream is 0, 1, 2.
make another IntStream for each integer in the outer stream from 0 to number of columns - also 3.
You cannot increment v in a stream because it must be "effectively final". Instead we use the equation board[j][i] = 1 + (j + m * i) which is effectively the similar to computing the index of the value if you were to flatten board into a single array (1D matrix).
import java.util.stream.IntStream;
import java.util.Arrays;
class J {
public static void main(String[] args) {
int n = 4;
int m = 2;
// n x m matrix
int[][] board = new int[n][m];
// Cols (m)
IntStream.range(0, m).forEach( i -> {
// Rows (n)
IntStream.range(0, n).forEach( j -> {
// Multiply i by row count
// to account for the row we are in
board[j][i] = 1 + (j + n * i);
});
});
System.out.println(Arrays.deepToString(board));
}
}
Output:
[[1, 5], [2, 6], [3, 7], [4, 8]]
Note: Just because streams allow you to write a neater, functional programming-like syntax, there is often times an associated performance toll. It stresses the idea, "why fix what's not broken?". For a 3 x 3 board, you probably won't see a difference. But, if your board were a lot larger, it probably won't prove itself to be worth it, considering all the objects created and extra space used behind the scene. Sometimes a simple for-loop (especially when working with arrays) is better.
Remember, simplicity is key.
Here is a stream solution that is "nested for cycles" look a like.
Get the sequence of the numbers from 0 to N(Upper limit exclusive -> N = board.length;
Get the sequence of the numbers from 0 to M(Upper limit exclusive -> M = board[i].length;
For each of the couples (i, j) set the value of board[i][j] using the func (i + j * step + 1) where step is defined as 5.
Output for int[][] board = new int[4][2];
[[1, 6], [2, 7], [3, 8], [4, 9]]
Output for int[][] board = new int[4][3];
[[1, 6, 11], [2, 7, 12], [3, 8, 13], [4, 9, 14]]
int step = 5;
IntStream.range(0, board.length).forEach(i ->
IntStream.range(0, board[i].length).forEach(j ->
board[i][j] = i + j * step + 1
)
);
You can use a combination of the IntStream methods range and iterate:
int width = 4, height = 2;
IntStream.rangeClosed(1, width)
.mapToObj(column -> IntStream.iterate(column, v -> v + width)
.limit(height)
.toArray())
.toArray(int[][]::new);
By using iterate for the inner loop and rangeClosed(1, width) instead of range(0, width) we can simplify the calcuation a bit. Using toArray removes the need to create and modify the array yourself and would enable parallel processing.
It is important to actually use streams properly to make them more than just weird looking for loops. Merely replacing your loop by a range(0, x).forEach(...) to modify a local variable does not really "use streams", and it is better to stick to for loops then.
Related
I'm working on the following task.
Given an array of n integers and two integer numbers m and k.
You can add any positive integer to any element of the array such that
the total value does not exceed k.
The task is to maximize the
multiples of m in the resultant array.
Consider the following example.
Input:
n = 5, m = 2, k = 2, arr[] = [1, 2, 3, 4, 5]
Let's add 1 to the element arr[0] and 1 to arr[2] then the final array would be:
[2, 2, 4, 4, 5]
Now there are four (4) elements which are multiples of m (2).
I am not getting correct output.
My code:
public class Main {
public static void main(String[] args) {
int n = 5;
int m = 4;
int k = 3;
int count = 0;
int[] arr = {17, 8, 9, 1, 4};
for (int i = 0; i < n; i++) {
for (int j = 0; j <= k; j++) {
// check initial
if (arr[i] % m == 0) {
break;
}
// add
arr[i] = arr[i] + j;
// check again
if (arr[i] % m == 0) {
count++;
break;
}
}
}
System.out.println("Final Array : " + Arrays.toString(arr));
System.out.println("Count : " + count);
}
}
This task boils down to a well-known Dynamic programming algorithm called Knapsack problem after a couple of simple manipulations with the given array.
This approach doesn't require sorting and would be advantages when k is much smaller n.
We can address the problem in the following steps:
Iterate over the given array and count all the numbers that are already divisible by m (this number is stored in the variable count in the code below).
While iterating, for every element of the array calculate the difference between m and remainder from the division of this element by m. Which would be equal to m - currentElement % m. If the difference is smaller or equal to k (it can cave this difference) it should be added to the list (differences in the code below) and also accumulated in a variable which is meant to store the total difference (totalDiff). All the elements which produce difference that exceeds k would be omitted.
If the total difference is less than or equal to k - we are done, the return value would be equal to the number of elements divisible by m plus the size of the list of differences.
Otherwise, we need to apply the logic of the Knapsack problem to the list of differences.
The idea behind the method getBestCount() (which is an implementation Knapsack problem) boils down to generating the "2D" array (a nested array of length equal to the size of the list of differences +1, in which every inner array having the length of k+1) and populating it with maximum values that could be achieved for various states of the Knapsack.
Each element of this array would represent the maximum total number of elements which can be adjusted to make them divisible by m for the various sizes of the Knapsack, i.e. number of items available from the list of differences, and different number of k (in the range from 0 to k inclusive).
The best way to understand how the algorithm works is to draw a table on a piece of paper and fill it with numbers manually (follow the comments in the code, some intermediate variables were introduced only for the purpose of making it easier to grasp, and also see the Wiki article linked above).
For instance, if the given array is [1, 8, 3, 9, 5], k=3 and m=3. We can see 2 elements divisible by m - 3 and 9. Numbers 1, 8, 5 would give the following list of differences [2, 1, 1]. Applying the logic of the Knapsack algorithm, we should get the following table:
[0, 0, 0, 0]
[0, 0, 1, 1]
[0, 1, 1, 2]
[0, 1, 2, 2]
We are interested in the value right most column of the last row, which is 2 plus 2 (number of elements divisible by 3) would give us 4.
Note: that code provided below can dial only with positive numbers. I don't want to shift the focus from the algorithm to such minor details. If OP or reader of the post are interested in making the code capable to work with negative number as well, I'm living the task of adjusting the code for them as an exercise. Hint: only a small change in the countMultiplesOfM() required for that purpose.
That how it might be implemented:
public static int countMultiplesOfM(int[] arr, int k, int m) {
List<Integer> differences = new ArrayList<>();
int count = 0;
long totalDiff = 0; // counter for the early kill - case when `k >= totalDiff`
for (int next : arr) {
if (next % m == 0)
count++; // number is already divisible by `m` we can increment the count and from that moment we are no longer interested in it
else if (m - next % m <= k) {
differences.add(m - next % m);
totalDiff += m - next % m;
}
}
if (totalDiff <= k) { // early kill - `k` is large enough to adjust all numbers in the `differences` list
return count + differences.size();
}
return count + getBestCount(differences, k); // fire the rest logic
}
// Knapsack Algorithm implementation
public static int getBestCount(List<Integer> differences, int knapsackSize) {
int[][] tab = new int[differences.size() + 1][knapsackSize + 1];
for (int numItemAvailable = 1; numItemAvailable < tab.length; numItemAvailable++) {
int next = differences.get(numItemAvailable - 1); // next available item which we're trying to place to knapsack to Maximize the current total
for (int size = 1; size < tab[numItemAvailable].length; size++) {
int prevColMax = tab[numItemAvailable][size - 1]; // maximum result for the current size - 1 in the current row of the table
int prevRowMax = tab[numItemAvailable - 1][size]; // previous maximum result for the current knapsack's size
if (next <= size) { // if it's possible to fit the next item in the knapsack
int prevRowMaxWithRoomForNewItem = tab[numItemAvailable - 1][size - next] + 1; // maximum result from the previous row for the size = `current size - next` (i.e. the closest knapsack size which guarantees that there would be a space for the new item)
tab[numItemAvailable][size] = Math.max(prevColMax, prevRowMaxWithRoomForNewItem);
} else {
tab[numItemAvailable][size] = Math.max(prevRowMax, prevColMax); // either a value in the previous row or a value in the previous column of the current row
}
}
}
return tab[differences.size()][knapsackSize];
}
main()
public static void main(String[] args) {
System.out.println(countMultiplesOfM(new int[]{17, 8, 9, 1, 4}, 3, 4));
System.out.println(countMultiplesOfM(new int[]{1, 2, 3, 4, 5}, 2, 2));
System.out.println(countMultiplesOfM(new int[]{1, 8, 3, 9, 5}, 3, 3));
}
Output:
3 // input array [17, 8, 9, 1, 4], m = 4, k = 3
4 // input array [1, 2, 3, 4, 5], m = 2, k = 2
4 // input array [1, 8, 3, 9, 5], m = 3, k = 3
A link to Online Demo
You must change 2 line in your code :
if(arr[i]%m==0)
{
count++; // add this line
break;
}
// add
arr[i]=arr[i]+1; // change j to 1
// check again
if(arr[i]%m==0)
{
count++;
break;
}
The first is because the number itself is divisible.
and The second is because you add a number to it each time.That is wrong.
for example chenge your arr to :
int[] arr ={17,8,10,2,4};
your output is :
Final Array : [20, 8, 16, 8, 4]
and That is wrong because 16-10=6 and is bigger than k=3.
I believe the problem is that you aren't processing the values in ascending order of the amount by which to adjust.
To solve this I started by using a stream to preprocess the array. This could be done using other methods.
map the values to the amount to make each one, when added, divisible by m.
filter out those that equal to m' (already divisible by m`)
sort in ascending order.
Once that is done. Intialize max to the difference between the original array length and the processed length. This is the number already divisible by m.
As the list is iterated
check to see if k > amount needed. If so, subtract from k and increment max
otherwise break out of the loop (because of the sort, no value remaining can be less than k)
public static int maxDivisors(int m, int k, int[] arr) {
int[] need = Arrays.stream(arr).map(v -> m - v % m)
.filter(v -> v != m).sorted().toArray();
int max = arr.length - need.length;
for (int val : need) {
if (k >= val) {
k -= val;
max++;
} else {
break;
}
}
return max;
}
int m = 4;
int k = 3;
int[] arr ={17,8,9,1,4};
int count = maxDivisors(m, k, arr);
System.out.println(count);
prints
3
I have a piece of Matlab code but could not find out how to convert to a Java code. What does this row mean? How can be converted to Java?
Matlab code:
b = all(x <= y) && any(x<y);
Hint: x = [1,2,3,4,5], y = [5,4,3,2,1]
What is b as the result?
You can use Java Streams like in
var x = new int[] {1, 2, 3, 4, 5};
var y = new int[] {5, 4, 3, 2, 1};
var b = IntStream.range(0, x.length).allMatch(i -> x[i] <= y[i])
&& IntStream.range(0, x.length).anyMatch(i -> x[i] < y[i]);
not handling x and y having different sizes!
From the matlab official site:
A <= B returns a logical array with elements set to logical 1 (true) where A is less than or equal to B; otherwise, the element is logical 0 (false).
Up to this you have 11100 in the first comparison.
B = all(A) tests along the first array dimension of A whose size does not equal 1, and determines if the elements are all nonzero or logical 1 (true). In practice, all is a natural extension of the logical AND operator.
So all(x <= y) must return 0.
B = any(A) tests along the first array dimension of A whose size does not equal 1, and determines if any element is a nonzero number or logical 1 (true). In practice, any is a natural extension of the logical OR operator.
So any(x < y) should be 1.
&& is AND
The rest is up to you
I have an array of numbers and another number K.
My task is to reduce the number of distinct elements in the array. For that, I can update the array several times. For updating the array, I have to follow these steps:
Select an element at index i and add that element by K, and reduce all other remaining elements by K.
For updating an array I can select the same index several times.
Example:
K = 1
Array: [3,1,3]
Answer: 3
I am picking index = 1, as [3-1, 1+1, 3-1] = [2,2,2] so we have number 2 that appears 3 times so this element occurs maximum number of times. So answer is 3.
Another example:
K = 1
Array: [1,2,2]
Answer: 2
It's not possible to make all elements same, so we have number 2 that appears 2 times, so answer is 2.
Array size can be [1, 1000], and the value of K and elements in array is in range [0, 1000]
Here is my code that I tried, my my approach is not correct.
public static int process(int K, int[] A) {
Map<Integer, Integer> map = new TreeMap<>();
for (int key : A) {
map.put(key, map.getOrDefault(key, 0) + 1);
}
int result = 0;
boolean flag = false;
int last = -1, cur = -1;
for (int key : map.keySet()) {
if (flag == false) {
flag = true;
last = key;
continue;
}
cur = key;
int a = map.get(last), b = map.get(cur);
if (Math.abs(last - cur) > K) {
result += a + b;
} else {
result += Math.max(a, b);
}
}
last = cur;
return result;
}
When looking at the examples with K = 1, it is clear that the answer
depends on the parity of the elements. Only elements with same parity can be set to the same level,
and all elements with same parity can be joined.
For example:
[2 4 6] -> [1 5 5] -> [2 4 4] -> [3 3 3]
[1 2 2] -> [2 1 1] ... no progress
With K = 1, we have to consider value modulo 2, i.e. modulo 2*K.
When K is different of one, for example K = 2, two numbers can be joined only there are separated by a distance multiple of 4, i.e. of 2*K.
[2 6 6] -> [4 4 4]
For K different from 1, instead of creating buckets for numbers with same parity,
we just create buckets according to value modulo 2K.
We just have to pay attention to use the modulo and not the remainder, the values are different for negative values.
Then the answer if simply the highest size of a bucket.
Output:
K = 1 Array : 3 1 3 -> 3
K = 1 Array : 1 2 2 -> 2
K = 1 Array : 2 3 4 7 4 9 11 -> 4
K = 1 Array : -3 -1 2 3 -> 3
K = 3 Array : -7 -1 0 1 2 4 5 -> 3
Here is a simple code in C++ to illustrate the algorithm.
In this code, the value val_modulo modulo 2K of each element is calculated.
Then, the orresponding counter is increased
Bucket[val_modulo] = Bucket[val_modulo] + 1
At the end, the highest value corresponds to the number of repetitions of the most repeated final value.
We may note that the number of non empty bucket corrresponds to the number of different
final values (not used in this code).
#include <iostream>
#include <vector>
#include <string>
#include <map>
void print (const std::vector<int> &A, const std::string &after = "\n", const std::string &before = "") {
std::cout << before;
for (int x: A) {
std::cout << x << " ";
}
std::cout << after;
}
int Modulo (int n, int mod) {
int ans = n % mod;
if (ans < 0) ans += mod;
return ans;
}
int max_equal(int K, std::vector<int> A) {
K = std::abs(K); // useful befoe taking the modulo
std::map<int, int> Buckets;
int nmax = 0;
int mod = 2*K;
for (int x: A) {
int val_modulo = Modulo (x, mod); // and not x*mod, as x can be negative
Buckets[val_modulo]++;
}
for (auto x: Buckets) {
if (x.second > nmax) {
nmax = x.second;
}
}
return nmax;
}
int main() {
std::vector<std::vector<int>> examples = {
{3, 1, 3},
{1, 2, 2},
{2, 3, 4, 7, 4, 9, 11},
{-3, -1, 2, 3},
{-7, -1, 0, 1, 2, 4, 5}
};
std::vector<int> tab_K = {1, 1, 1, 1, 3};
for (int i = 0; i < examples.size(); ++i) {
std::cout << "K = " << tab_K[i] << " Array : ";
print (examples[i], " -> ");
auto ans = max_equal (tab_K[i], examples[i]);
std::cout << ans << "\n";
}
return 0;
}
The problem is conceptual, and translating it in somewhat computing code.
Let's look:
We pick a number (by index, which is irrelevant), and for all the occurrences we add K. All others we subtract K And then the number of same occurrences must be maximal.
The same occurrences can only grow when the picked number + K is equal to another number - K.
The data structure:
A map with the array numbers as key, and mapped to frequency (how often the number occurs in the array).
So:
pickedNumber.value + K = otherNumber.value - K
=> otherNumber.value = pickedNumber.value + 2*K
Note that as there is only one single otherNumber, derived from the pickedNumber.
(It might occur more than once.)
And we want maximal:
pickedNumber.frequency + otherNumber.frequency maximal.
Though map is not really needed, a sorted array would do too, let's do a map:
The algorithm:
Kept simple.
int[] numbers = {3, 1, 3};
int index = pickedIndexOfBestSolution(numbers, 1);
System.out.println("Index: " + index);
int pickedIndexOfBestSolution(int[] numbers, int k) {
Map<Integer, Long> frequencyTable = IntStream.of(numbers)
.boxed()
.collect(Collectors.groupingBy(Function.identity(),
Collectors.counting()));
int bestNumber = frequencyTable.entrySet().stream()
.sorted(Comparator.comparingLong(e -> -e.getValue()
- frequencyTable.getOrDefault(e.getKey() + 2*k, 0L)))
.findFirst()
.map(e -> e.getKey())
.orElseThrow();
int index = -1;
while (numbers[++index] != bestNumber) {
}
return index;
}
The frequency table I filled using an IntStream and groupingBy (just as SQL).
As counting is done with long, I just kept that.
To find the max I counted the new occurrence count trying to add the "other" number's frequency too; 0 when no other number.
Instead of using .reverse() to reverse the comparison (largest, max, first), I took the negative value, which to me seems more calculatory.
Notice that a Stream with findFirst to find the max is probably optimal too: no need that the stream creates an intermediate list.
For the index I used brute force (while loop), a kind of indexOf.
Notice if there is no other number, it returns the index of a number with the most occurrences. Which is fine.
In short:
You see the different approach. Actually simpler, and more solid. In fact applying
some (minor) intelligence first. Trying to nail down the problem first.
Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 7 years ago.
Improve this question
Right now I'm trying to write a function that takes an array and an integer n, and gives a list of each size n combination (so a list of int arrays). I am able to write it using n nested loops, but this only works for a specific size of subset. I can't figure out how to generalize it to work for any size of combination. I think I need to use recursion?
This is the code for all combinations of 3 elements, and I need an algorithm for any number of elements.
import java.util.List;
import java.util.ArrayList;
public class combinatorics{
public static void main(String[] args) {
List<int[]> list = new ArrayList<int[]>();
int[] arr = {1,2,3,4,5};
combinations3(arr,list);
listToString(list);
}
static void combinations3(int[] arr, List<int[]> list){
for(int i = 0; i<arr.length-2; i++)
for(int j = i+1; j<arr.length-1; j++)
for(int k = j+1; k<arr.length; k++)
list.add(new int[]{arr[i],arr[j],arr[k]});
}
private static void listToString(List<int[]> list){
for(int i = 0; i<list.size(); i++){ //iterate through list
for(int j : list.get(i)){ //iterate through array
System.out.printf("%d ",j);
}
System.out.print("\n");
}
}
}
This is a well-studied problem of generating all k-subsets, or k-combinations, which can be easily done without recursion.
The idea is to have array of size k keeping sequence of indices of elements from the input array (which are numbers from 0 to n - 1) in increasing order. (Subset then can be created by taking items by these indices from the initial array.) So we need to generate all such index sequences.
First index sequence will be [0, 1, 2, ... , k - 1], on the second step it switches to [0, 1, 2,..., k], then to [0, 1, 2, ... k + 1] and so on. The last possible sequence will be [n - k, n - k + 1, ..., n - 1].
On each step, algorithm looks for the closest to the end item which can be incremented, increments it and fills up items right to that item.
To illustrate, consider n = 7 and k = 3. First index sequence is [0, 1, 2], then [0, 1, 3] and so on... At some point we have [0, 5, 6]:
[0, 5, 6] <-- scan from the end: "6" cannot be incremented, "5" also, but "0" can be
[1, ?, ?] <-- "0" -> "1"
[1, 2, 3] <-- fill up remaining elements
next iteration:
[1, 2, 3] <-- "3" can be incremented
[1, 2, 4] <-- "3" -> "4"
Thus, [0, 5, 6] is followed by [1, 2, 3], then goes [1, 2, 4] etc.
Code:
int[] input = {10, 20, 30, 40, 50}; // input array
int k = 3; // sequence length
List<int[]> subsets = new ArrayList<>();
int[] s = new int[k]; // here we'll keep indices
// pointing to elements in input array
if (k <= input.length) {
// first index sequence: 0, 1, 2, ...
for (int i = 0; (s[i] = i) < k - 1; i++);
subsets.add(getSubset(input, s));
for(;;) {
int i;
// find position of item that can be incremented
for (i = k - 1; i >= 0 && s[i] == input.length - k + i; i--);
if (i < 0) {
break;
}
s[i]++; // increment this item
for (++i; i < k; i++) { // fill up remaining items
s[i] = s[i - 1] + 1;
}
subsets.add(getSubset(input, s));
}
}
// generate actual subset by index sequence
int[] getSubset(int[] input, int[] subset) {
int[] result = new int[subset.length];
for (int i = 0; i < subset.length; i++)
result[i] = input[subset[i]];
return result;
}
If I understood your problem correctly, this article seems to point to what you're trying to do.
To quote from the article:
Method 1 (Fix Elements and Recur)
We create a temporary array ‘data[]’ which stores all outputs one by
one. The idea is to start from first index (index = 0) in data[], one
by one fix elements at this index and recur for remaining indexes. Let
the input array be {1, 2, 3, 4, 5} and r be 3. We first fix 1 at index
0 in data[], then recur for remaining indexes, then we fix 2 at index
0 and recur. Finally, we fix 3 and recur for remaining indexes. When
number of elements in data[] becomes equal to r (size of a
combination), we print data[].
Method 2 (Include and Exclude every element)
Like the above method, We create a temporary array data[]. The idea
here is similar to Subset Sum Problem. We one by one consider every
element of input array, and recur for two cases:
The element is included in current combination (We put the element in data[] and increment next available index in data[])
The element is excluded in current combination (We do not put the element and do not change index)
When number of elements in data[] become equal to r (size of a
combination), we print it.
I am stuck with one problem :
Finding all the possible ways of dividing a group of size 'n' into subgroups of size 'k'. (here n%k =0)
e.g., let set be {1,2,3,4,5,6} to divide into subgroups of 3 (k = 3,n = 6), possible sets are
a) {1,2,3},{4,5,6}
b) {1,3,5},{2,4,6}
c) {1,3,6},{2,4,5}
d) {1,3,4},{2,5,6} etc....
What i tried doing was, to first find all the combination of size k from the set.
Then loop through these combinations and find out which all combinations can be grouped together to find the list of subgroups.
But I believe the time complexity is pretty bad in this approach. Is there any better approach for this problem?
I would use a recursive method. I think this one has the optimal running time, since it exactly produces all needed subsets.
public static void solve(int[] a, int k, int i, List<List<Integer>> subsets) {
if (i == a.length) {
for (List<Integer> subset : subsets) {
System.out.print(subset);
}
System.out.println();
} else {
// loop over all subsets and try to put a[i] in
for (int j = 0; j < subsets.size(); j++) {
if (subsets.get(j).size() < k) {
// subset j not full
subsets.get(j).add(a[i]);
solve(a, k, i+1, subsets); // do recursion
subsets.get(j).remove((Integer)a[i]);
if (subsets.get(j).size() == 0) {
// don't skip empty subsets, so you won't get duplicates
break;
}
}
}
}
}
Usage:
public static void main(String[] args) {
int[] a = {1, 2, 3, 4, 5, 6};
int k = 3;
List<List<Integer>> subsets = new ArrayList<List<Integer>>(a.length / k);
for (int i = 0; i < a.length / k; i++)
subsets.add(new ArrayList<Integer>(k));
solve(a, k, 0, subsets);
}
Prints:
[1, 2, 3][4, 5, 6]
[1, 2, 4][3, 5, 6]
[1, 2, 5][3, 4, 6]
[1, 2, 6][3, 4, 5]
[1, 3, 4][2, 5, 6]
[1, 3, 5][2, 4, 6]
[1, 3, 6][2, 4, 5]
[1, 4, 5][2, 3, 6]
[1, 4, 6][2, 3, 5]
[1, 5, 6][2, 3, 4]
Think about it combinatorially. If n % k != 0, you can't do it because you'll end up with one set that has less than k elements, so start off with checking whether that is the case.
Afterwards, all you need to do is recursively produce k-combinations from an n-i*k set for all i in [0; n/k]. Algorithms for producing all k-combinations of a given set can be found on SO easy enough. The idea is: there are (n choose k) possible such sets you can choose for your first set; from the remaining n-k elements, you can choose ((n-k) choose k) sets); from the remaining n-2k elements, you can choose ((n-2k) choose k) sets and so on. Assuming the order of your sets doesn't matter, you have (n choose k) * ((n-k) choose k) * ... * ((n-(n-1)k) choose k) / ((n/k)!) possibilities to choose your sets, which depending on k can be exponential in the number of elements your original set has, so if you really want to produce each and every one of them, you won't get below exponential complexity.
I found a formula for the # of possible subgroupings, in case anyone finds it interesting. (Would this be considered too off-topic? Am I posting this correctly?)
First let m = n/k be the # of subgroups. Now let the first subgroup be fixed as the first k elements of the group, the second subgroup be the next k, and so on. If we consider all possible permutations of the group, this will give us all the different subgroupings. There are n! permutations of the n elements, but we don't care about ordering, so we factor out the k! permutations of each of the m subgroups and the m! permutations of the subgroups themselves. This gives us:
n!/(m!*(k!)^m).
As a check, if k = 1 or k = n, this gives us 1 subgrouping. In the original example, n = 6, k = 3, m = 2, and we get 10 possible subgroupings (which Heuster's code found).
Now, if you compare this expression to the one G. Bach gave and use (n choose k) = n!/(k!*(n-k)!), you will see that all the (n-k)! terms cancel and it reduces to the expression above.
Bonus: if you use Stirling's approximation for n!, then the expression simplifies nicely, and you get that the # of subgroupings scales as (m^n)/m!.