Permutation of array

Question

For example I have this array:

int a[] = new int[]{3,4,6,2,1};

I need list of all permutations such that if one is like this, {3,2,1

Answer 1

Example with primitive array:

public static void permute(int[] intArray, int start) {
    for(int i = start; i < intArray.length; i++){
        int temp = intArray[start];
        intArray[start] = intArray[i];
        intArray[i] = temp;
        permute(intArray, start + 1);
        intArray[i] = intArray[start];
        intArray[start] = temp;
    }
    if (start == intArray.length - 1) {
        System.out.println(java.util.Arrays.toString(intArray));
    }
}

public static void main(String[] args){
    int intArr[] = {1, 2, 3};
    permute(intArr, 0);
}

Answer 2

Here is how you can print all permutations in 10 lines of code:

public class Permute{
    static void permute(java.util.List<Integer> arr, int k){
        for(int i = k; i < arr.size(); i++){
            java.util.Collections.swap(arr, i, k);
            permute(arr, k+1);
            java.util.Collections.swap(arr, k, i);
        }
        if (k == arr.size() -1){
            System.out.println(java.util.Arrays.toString(arr.toArray()));
        }
    }
    public static void main(String[] args){
        Permute.permute(java.util.Arrays.asList(3,4,6,2,1), 0);
    }
}

You take first element of an array (k=0) and exchange it with any element (i) of the array. Then you recursively apply permutation on array starting with second element. This way you get all permutations starting with i-th element. The tricky part is that after recursive call you must swap i-th element with first element back, otherwise you could get repeated values at the first spot. By swapping it back we restore order of elements (basically you do backtracking).

Iterators and Extension to the case of repeated values

The drawback of previous algorithm is that it is recursive, and does not play nicely with iterators. Another issue is that if you allow repeated elements in your input, then it won't work as is.

For example, given input [3,3,4,4] all possible permutations (without repetitions) are

[3, 3, 4, 4]
[3, 4, 3, 4]
[3, 4, 4, 3]
[4, 3, 3, 4]
[4, 3, 4, 3]
[4, 4, 3, 3]

(if you simply apply permute function from above you will get [3,3,4,4] four times, and this is not what you naturally want to see in this case; and the number of such permutations is 4!/(2!*2!)=6)

It is possible to modify the above algorithm to handle this case, but it won't look nice. Luckily, there is a better algorithm (I found it here) which handles repeated values and is not recursive.

First note, that permutation of array of any objects can be reduced to permutations of integers by enumerating them in any order.

To get permutations of an integer array, you start with an array sorted in ascending order. You 'goal' is to make it descending. To generate next permutation you are trying to find the first index from the bottom where sequence fails to be descending, and improves value in that index while switching order of the rest of the tail from descending to ascending in this case.

Here is the core of the algorithm:

//ind is an array of integers
for(int tail = ind.length - 1;tail > 0;tail--){
    if (ind[tail - 1] < ind[tail]){//still increasing

        //find last element which does not exceed ind[tail-1]
        int s = ind.length - 1;
        while(ind[tail-1] >= ind[s])
            s--;

        swap(ind, tail-1, s);

        //reverse order of elements in the tail
        for(int i = tail, j = ind.length - 1; i < j; i++, j--){
            swap(ind, i, j);
        }
        break;
    }
}

Here is the full code of iterator. Constructor accepts an array of objects, and maps them into an array of integers using HashMap.

import java.lang.reflect.Array;
import java.util.*;
class Permutations<E> implements  Iterator<E[]>{

    private E[] arr;
    private int[] ind;
    private boolean has_next;

    public E[] output;//next() returns this array, make it public

    Permutations(E[] arr){
        this.arr = arr.clone();
        ind = new int[arr.length];
        //convert an array of any elements into array of integers - first occurrence is used to enumerate
        Map<E, Integer> hm = new HashMap<E, Integer>();
        for(int i = 0; i < arr.length; i++){
            Integer n = hm.get(arr[i]);
            if (n == null){
                hm.put(arr[i], i);
                n = i;
            }
            ind[i] = n.intValue();
        }
        Arrays.sort(ind);//start with ascending sequence of integers


        //output = new E[arr.length]; <-- cannot do in Java with generics, so use reflection
        output = (E[]) Array.newInstance(arr.getClass().getComponentType(), arr.length);
        has_next = true;
    }

    public boolean hasNext() {
        return has_next;
    }

    /**
     * Computes next permutations. Same array instance is returned every time!
     * @return
     */
    public E[] next() {
        if (!has_next)
            throw new NoSuchElementException();

        for(int i = 0; i < ind.length; i++){
            output[i] = arr[ind[i]];
        }


        //get next permutation
        has_next = false;
        for(int tail = ind.length - 1;tail > 0;tail--){
            if (ind[tail - 1] < ind[tail]){//still increasing

                //find last element which does not exceed ind[tail-1]
                int s = ind.length - 1;
                while(ind[tail-1] >= ind[s])
                    s--;

                swap(ind, tail-1, s);

                //reverse order of elements in the tail
                for(int i = tail, j = ind.length - 1; i < j; i++, j--){
                    swap(ind, i, j);
                }
                has_next = true;
                break;
            }

        }
        return output;
    }

    private void swap(int[] arr, int i, int j){
        int t = arr[i];
        arr[i] = arr[j];
        arr[j] = t;
    }

    public void remove() {

    }
}

Usage/test:

    TCMath.Permutations<Integer> perm = new TCMath.Permutations<Integer>(new Integer[]{3,3,4,4,4,5,5});
    int count = 0;
    while(perm.hasNext()){
        System.out.println(Arrays.toString(perm.next()));
        count++;
    }
    System.out.println("total: " + count);

Prints out all 7!/(2!*3!*2!)=210 permutations.

Answer 3

Here is an implementation of the Permutation in Java:

Permutation - Java

You should have a check on it!

Edit: code pasted below to protect against link-death:

// Permute.java -- A class generating all permutations

import java.util.Iterator;
import java.util.NoSuchElementException;
import java.lang.reflect.Array;

public class Permute implements Iterator {

   private final int size;
   private final Object [] elements;  // copy of original 0 .. size-1
   private final Object ar;           // array for output,  0 .. size-1
   private final int [] permutation;  // perm of nums 1..size, perm[0]=0

   private boolean next = true;

   // int[], double[] array won't work :-(
   public Permute (Object [] e) {
      size = e.length;
      elements = new Object [size];    // not suitable for primitives
      System.arraycopy (e, 0, elements, 0, size);
      ar = Array.newInstance (e.getClass().getComponentType(), size);
      System.arraycopy (e, 0, ar, 0, size);
      permutation = new int [size+1];
      for (int i=0; i<size+1; i++) {
         permutation [i]=i;
      }
   }

   private void formNextPermutation () {
      for (int i=0; i<size; i++) {
         // i+1 because perm[0] always = 0
         // perm[]-1 because the numbers 1..size are being permuted
         Array.set (ar, i, elements[permutation[i+1]-1]);
      }
   }

   public boolean hasNext() {
      return next;
   }

   public void remove() throws UnsupportedOperationException {
      throw new UnsupportedOperationException();
   }

   private void swap (final int i, final int j) {
      final int x = permutation[i];
      permutation[i] = permutation [j];
      permutation[j] = x;
   }

   // does not throw NoSuchElement; it wraps around!
   public Object next() throws NoSuchElementException {

      formNextPermutation ();  // copy original elements

      int i = size-1;
      while (permutation[i]>permutation[i+1]) i--;

      if (i==0) {
         next = false;
         for (int j=0; j<size+1; j++) {
            permutation [j]=j;
         }
         return ar;
      }

      int j = size;

      while (permutation[i]>permutation[j]) j--;
      swap (i,j);
      int r = size;
      int s = i+1;
      while (r>s) { swap(r,s); r--; s++; }

      return ar;
   }

   public String toString () {
      final int n = Array.getLength(ar);
      final StringBuffer sb = new StringBuffer ("[");
      for (int j=0; j<n; j++) {
         sb.append (Array.get(ar,j).toString());
         if (j<n-1) sb.append (",");
      }
      sb.append("]");
      return new String (sb);
   }

   public static void main (String [] args) {
      for (Iterator i = new Permute(args); i.hasNext(); ) {
         final String [] a = (String []) i.next();
         System.out.println (i);
      }
   }
}

Answer 4

Do like this...

import java.util.ArrayList;
import java.util.Arrays;

public class rohit {

    public static void main(String[] args) {
        ArrayList<Integer> a=new ArrayList<Integer>();
        ArrayList<Integer> b=new ArrayList<Integer>();
        b.add(1);
        b.add(2);
        b.add(3);
        permu(a,b);
    }

    public static void permu(ArrayList<Integer> prefix,ArrayList<Integer> value) {
        if(value.size()==0) {
            System.out.println(prefix);
        } else {
            for(int i=0;i<value.size();i++) {
                ArrayList<Integer> a=new ArrayList<Integer>();
                a.addAll(prefix);
                a.add(value.get(i));

                ArrayList<Integer> b=new ArrayList<Integer>();

                b.addAll(value.subList(0, i));
                b.addAll(value.subList(i+1, value.size()));

                permu(a,b);
            }
        }
    }

}

Answer 5

A simple java implementation, refer to c++ std::next_permutation:

public static void main(String[] args){
    int[] list = {1,2,3,4,5};
    List<List<Integer>> output = new Main().permute(list);
    for(List result: output){
        System.out.println(result);
    }

}

public List<List<Integer>> permute(int[] nums) {
    List<List<Integer>> list = new ArrayList<List<Integer>>();
    int size = factorial(nums.length);

    // add the original one to the list
    List<Integer> seq = new ArrayList<Integer>();
    for(int a:nums){
        seq.add(a);
    }
    list.add(seq);

    // generate the next and next permutation and add them to list
    for(int i = 0;i < size - 1;i++){
        seq = new ArrayList<Integer>();
        nextPermutation(nums);
        for(int a:nums){
            seq.add(a);
        }
        list.add(seq);
    }
    return list;
}


int factorial(int n){
    return (n==1)?1:n*factorial(n-1);
}

void nextPermutation(int[] nums){
    int i = nums.length -1; // start from the end

    while(i > 0 && nums[i-1] >= nums[i]){
        i--;
    }

    if(i==0){
        reverse(nums,0,nums.length -1 );
    }else{

        // found the first one not in order 
        int j = i;

        // found just bigger one
        while(j < nums.length && nums[j] > nums[i-1]){
            j++;
        }
        //swap(nums[i-1],nums[j-1]);
        int tmp = nums[i-1];
        nums[i-1] = nums[j-1];
        nums[j-1] = tmp;
        reverse(nums,i,nums.length-1);  
    }
}

// reverse the sequence
void reverse(int[] arr,int start, int end){
    int tmp;
    for(int i = 0; i <= (end - start)/2; i++ ){
        tmp = arr[start + i];
        arr[start + i] = arr[end - i];
        arr[end - i ] = tmp;
    }
}

Answer 6

There are n! total permutations for the given array size n. Here is code written in Java using DFS.

public List<List<Integer>> permute(int[] nums) {
    List<List<Integer>> results = new ArrayList<List<Integer>>();
    if (nums == null || nums.length == 0) {
        return results;
    }
    List<Integer> result = new ArrayList<>();
    dfs(nums, results, result);
    return results;
}

public void dfs(int[] nums, List<List<Integer>> results, List<Integer> result) {
    if (nums.length == result.size()) {
        List<Integer> temp = new ArrayList<>(result);
        results.add(temp);
    }        
    for (int i=0; i<nums.length; i++) {
        if (!result.contains(nums[i])) {
            result.add(nums[i]);
            dfs(nums, results, result);
            result.remove(result.size() - 1);
        }
    }
}

For input array [3,2,1,4,6], there are totally 5! = 120 possible permutations which are:

[[3,4,6,2,1],[3,4,6,1,2],[3,4,2,6,1],[3,4,2,1,6],[3,4,1,6,2],[3,4,1,2,6],[3,6,4,2,1],[3,6,4,1,2],[3,6,2,4,1],[3,6,2,1,4],[3,6,1,4,2],[3,6,1,2,4],[3,2,4,6,1],[3,2,4,1,6],[3,2,6,4,1],[3,2,6,1,4],[3,2,1,4,6],[3,2,1,6,4],[3,1,4,6,2],[3,1,4,2,6],[3,1,6,4,2],[3,1,6,2,4],[3,1,2,4,6],[3,1,2,6,4],[4,3,6,2,1],[4,3,6,1,2],[4,3,2,6,1],[4,3,2,1,6],[4,3,1,6,2],[4,3,1,2,6],[4,6,3,2,1],[4,6,3,1,2],[4,6,2,3,1],[4,6,2,1,3],[4,6,1,3,2],[4,6,1,2,3],[4,2,3,6,1],[4,2,3,1,6],[4,2,6,3,1],[4,2,6,1,3],[4,2,1,3,6],[4,2,1,6,3],[4,1,3,6,2],[4,1,3,2,6],[4,1,6,3,2],[4,1,6,2,3],[4,1,2,3,6],[4,1,2,6,3],[6,3,4,2,1],[6,3,4,1,2],[6,3,2,4,1],[6,3,2,1,4],[6,3,1,4,2],[6,3,1,2,4],[6,4,3,2,1],[6,4,3,1,2],[6,4,2,3,1],[6,4,2,1,3],[6,4,1,3,2],[6,4,1,2,3],[6,2,3,4,1],[6,2,3,1,4],[6,2,4,3,1],[6,2,4,1,3],[6,2,1,3,4],[6,2,1,4,3],[6,1,3,4,2],[6,1,3,2,4],[6,1,4,3,2],[6,1,4,2,3],[6,1,2,3,4],[6,1,2,4,3],[2,3,4,6,1],[2,3,4,1,6],[2,3,6,4,1],[2,3,6,1,4],[2,3,1,4,6],[2,3,1,6,4],[2,4,3,6,1],[2,4,3,1,6],[2,4,6,3,1],[2,4,6,1,3],[2,4,1,3,6],[2,4,1,6,3],[2,6,3,4,1],[2,6,3,1,4],[2,6,4,3,1],[2,6,4,1,3],[2,6,1,3,4],[2,6,1,4,3],[2,1,3,4,6],[2,1,3,6,4],[2,1,4,3,6],[2,1,4,6,3],[2,1,6,3,4],[2,1,6,4,3],[1,3,4,6,2],[1,3,4,2,6],[1,3,6,4,2],[1,3,6,2,4],[1,3,2,4,6],[1,3,2,6,4],[1,4,3,6,2],[1,4,3,2,6],[1,4,6,3,2],[1,4,6,2,3],[1,4,2,3,6],[1,4,2,6,3],[1,6,3,4,2],[1,6,3,2,4],[1,6,4,3,2],[1,6,4,2,3],[1,6,2,3,4],[1,6,2,4,3],[1,2,3,4,6],[1,2,3,6,4],[1,2,4,3,6],[1,2,4,6,3],[1,2,6,3,4],[1,2,6,4,3]]

Hope this helps.