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, {3,2,1,4,6}, others must not be the same. I know that if the length of the array is n then there are n! possible combinations. How can this algorithm be written?

Update: thanks, but I need a pseudo code algorithm like:

for(int i=0;i<a.length;i++){
 // code here
}

Just algorithm. Yes, API functions are good, but it does not help me too much.

• Marcelo Cantos: There aren't 2^n possible combinations. There are n! permutations. Plus, I don't understand the question. Are you simply trying to exclude a single permutation, {3,2,1,4,6}?
• user466534: yes sorry n! no all permutation should be unique
• Bryan Rayner: Could other language tags be added to this? Since this is an algorithm, it would be good to have multiple implementations in various languages.

Answer 1

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.

• raam86: Great Answer. Can you please explain why it's 4!/(2!2!)=6 and not 4!/(2!)=12
• Yevgen Yampolskiy: First of all, I know that answer is 6 (from my [3,3,4,4] example). To derive the formula, think about [3,3,4,4] as two blue and two red balls. The question is how many ways to position balls (balls of the same color are the same). If you somehow position your balls, then interchanging of the blue balls (2! ways of doing that) or two red balls (2! ways of doing that) does not change anything. Now, we have 4! ways to place 4 balls, but permuting blue balls (2! ways) or red balls (2! ways) does not change positioning of the balls. So you get 4!/(2!*2!) as final answer
• Hengameh: Time complexity of first algorithm is O(n*n!), is that right?
• Anonymous: this is the fastest permutation algorithm i've tried. good job
• Dan Rosenstark: I rarely like long explanations on SO, but this is an awesome exception. Thanks for explaining!

Answer 2

If you're using C++, you can use std::next_permutation from the <algorithm> header file:

int a[] = {3,4,6,2,1};
int size = sizeof(a)/sizeof(a[0]);
std::sort(a, a+size);
do {
 // print a's elements
} while(std::next_permutation(a, a+size));

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);
 }
 }
}

• BlueRaja - Danny Pflughoeft: +1 please add the relevant code to your post though, in case the link ever goes down
• user124384: Thanks also for eliminating the line numbers. :P
• pandey_shubham: And the link went down. :)
• Some Guy: @BlueRaja-DannyPflughoeft Nice catch, the link did go down

Answer 4

According to wiki https://en.wikipedia.org/wiki/Heap%27s_algorithm

Heap's algorithm generates all possible permutations of n objects. It was first proposed by B. R. Heap in 1963. The algorithm minimizes movement: it generates each permutation from the previous one by interchanging a single pair of elements; the other n−2 elements are not disturbed. In a 1977 review of permutation-generating algorithms, Robert Sedgewick concluded that it was at that time the most effective algorithm for generating permutations by computer.

So if we want to do it in recursive manner, Sudo code is bellow.

procedure generate(n : integer, A : array of any):
 if n = 1 then
 output(A)
 else
 for i := 0; i < n - 1; i += 1 do
 generate(n - 1, A)
 if n is even then
 swap(A[i], A[n-1])
 else
 swap(A[0], A[n-1])
 end if
 end for
 generate(n - 1, A)
 end if

java code:

public static void printAllPermutations(
 int n, int[] elements, char delimiter) {
 if (n == 1) {
 printArray(elements, delimiter);
 } else {
 for (int i = 0; i < n - 1; i++) {
 printAllPermutations(n - 1, elements, delimiter);
 if (n % 2 == 0) {
 swap(elements, i, n - 1);
 } else {
 swap(elements, 0, n - 1);
 }
 }
 printAllPermutations(n - 1, elements, delimiter);
 }
}
private static void printArray(int[] input, char delimiter) {
 int i = 0;
 for (; i < input.length; i++) {
 System.out.print(input[i]);
 }
 System.out.print(delimiter);
}
private static void swap(int[] input, int a, int b) {
 int tmp = input[a];
 input[a] = input[b];
 input[b] = tmp;
}
public static void main(String[] args) {
 int[] input = new int[]{0,1,2,3};
 printAllPermutations(input.length, input, ',');
}

Answer 5

This a 2-permutation for a list wrapped in an iterator

import java.util.Iterator;
import java.util.LinkedList;
import java.util.List;
/* all permutations of two objects 
 * 
 * for ABC: AB AC BA BC CA CB
 * 
 * */
public class ListPermutation<T> implements Iterator {
 int index = 0;
 int current = 0;
 List<T> list;
 public ListPermutation(List<T> e) {
 list = e;
 }
 public boolean hasNext() {
 return !(index == list.size() - 1 && current == list.size() - 1);
 }
 public List<T> next() {
 if(current == index) {
 current++;
 }
 if (current == list.size()) {
 current = 0;
 index++;
 }
 List<T> output = new LinkedList<T>();
 output.add(list.get(index));
 output.add(list.get(current));
 current++;
 return output;
 }
 public void remove() {
 }
}

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.

Answer 7

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 8

Here is one using arrays and Java 8+

import java.util.Arrays;
import java.util.stream.IntStream;
public class HelloWorld {
 public static void main(String[] args) {
 int[] arr = {1, 2, 3, 5};
 permutation(arr, new int[]{});
 }
 static void permutation(int[] arr, int[] prefix) {
 if (arr.length == 0) {
 System.out.println(Arrays.toString(prefix));
 }
 for (int i = 0; i < arr.length; i++) {
 int i2 = i;
 int[] pre = IntStream.concat(Arrays.stream(prefix), IntStream.of(arr[i])).toArray();
 int[] post = IntStream.range(0, arr.length).filter(i1 -> i1 != i2).map(v -> arr[v]).toArray();
 permutation(post, pre);
 }
 }
}

Answer 9

Visual representation of the 3-item recursive solution: http://www.docdroid.net/ea0s/generatepermutations.pdf.html

Breakdown:

1. For a two-item array, there are two permutations:
   • The original array, and
   • The two elements swapped
2. For a three-item array, there are six permutations:
   • The permutations of the bottom two elements, then
   • Swap 1st and 2nd items, and the permutations of the bottom two element
   • Swap 1st and 3rd items, and the permutations of the bottom two elements.
   • Essentially, each of the items gets its chance at the first slot

Answer 10

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 11

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 12

Implementation via recursion (dynamic programming), in Java, with test case (TestNG).

---

Code

PrintPermutation.java

import java.util.Arrays;
/**
 * Print permutation of n elements.
 * 
 * @author eric
 * @date Oct 13, 2018 12:28:10 PM
 */
public class PrintPermutation {
 /**
 * Print permutation of array elements.
 * 
 * @param arr
 * @return count of permutation,
 */
 public static int permutation(int arr[]) {
 return permutation(arr, 0);
 }
 /**
 * Print permutation of part of array elements.
 * 
 * @param arr
 * @param n
 * start index in array,
 * @return count of permutation,
 */
 private static int permutation(int arr[], int n) {
 int counter = 0;
 for (int i = n; i < arr.length; i++) {
 swapArrEle(arr, i, n);
 counter += permutation(arr, n + 1);
 swapArrEle(arr, n, i);
 }
 if (n == arr.length - 1) {
 counter++;
 System.out.println(Arrays.toString(arr));
 }
 return counter;
 }
 /**
 * swap 2 elements in array,
 * 
 * @param arr
 * @param i
 * @param k
 */
 private static void swapArrEle(int arr[], int i, int k) {
 int tmp = arr[i];
 arr[i] = arr[k];
 arr[k] = tmp;
 }
}

PrintPermutationTest.java (test case via TestNG)

import org.testng.Assert;
import org.testng.annotations.Test;
/**
 * PrintPermutation test.
 * 
 * @author eric
 * @date Oct 14, 2018 3:02:23 AM
 */
public class PrintPermutationTest {
 @Test
 public void test() {
 int arr[] = new int[] { 0, 1, 2, 3 };
 Assert.assertEquals(PrintPermutation.permutation(arr), 24);
 int arrSingle[] = new int[] { 0 };
 Assert.assertEquals(PrintPermutation.permutation(arrSingle), 1);
 int arrEmpty[] = new int[] {};
 Assert.assertEquals(PrintPermutation.permutation(arrEmpty), 0);
 }
}

Answer 13

Instead of permutations, we can preferably call them combinations.

Irrespective of the coding language, we can use a simple approach, To append the array elements to the already existing list of combinations thus, using the dynamic programming approach.

This code focuses on those combinations without adjacency as well.

#include <iostream>
#include <vector>
using namespace std;
template <class myIterator, class T>
myIterator findDigit(myIterator first, myIterator last, T val)
{
 while(first != last) {
 if(*first == val) { break; }
 first++;
 }
 return first;
}
void printCombinations(vector<vector<int>> combinations)
{
 cout << "printing all " << combinations.size() << " combinations" << endl;
 for(int i=0; i<combinations.size(); i++) {
 cout << "[";
 for(int j=0; j<combinations[i].size(); j++) {
 cout << " " << combinations[i][j] << " ";
 }
 cout << "] , ";
 }
 return;
}
int main()
{
 vector<int> a = {1,2,3,4,5};
 vector<vector<int>> comb;
 vector<int> t;
 int len=a.size();
 for(int i=0; i<len; i++) {
 t.push_back(a.at(i));
 comb.push_back(t);
 t.clear();
 }
 for(int l=1; l<len; l++) {
 for(int j=0; j<comb.size(); j++) {
 if(comb[j].size()==l) {
 int t = comb[j].back();
 if(t != a.back()) {
 vector<int>::iterator it = findDigit(a.begin(), a.end(), t);
 for(std::vector<int>::iterator k=it+1; k!=a.end();k++) {
 vector<int> t (comb[j].begin(), comb[j].end());
 t.push_back(*k);
 comb.push_back(t);
 t.clear();
 }
 }
 }
 }
 }
 printCombinations(comb);
 return 0;
}

Although the complexity is a bit high, it's definitely lower than the recursion approach, especially when the array size is very large.

Output for the above array (or vecter, if you will) is :

printing all 31 combinations
[ 1 ], [ 2 ], [ 3 ], [ 4 ], [ 5 ], [ 1 2 ], [ 1 3 ], [ 1 4 ], [ 1 5 ], [ 2 3 ], [ 2 4 ], [ 2 5 ], [ 3 4 ], [ 3 5 ], [ 4 5 ], [ 1 
 2 3 ], [ 1 2 4 ], [ 1 2 5 ], [ 1 3 4 ], [ 1 3 5 ], [ 1 4 5 ], [ 2 3 4 ], [ 2 3 5 ], [ 2 4 5 ], [ 3 4 5 ], [ 1 2 3 4 
], [ 1 2 3 5 ], [ 1 2 4 5 ], [ 1 3 4 5 ], [ 2 3 4 5 ], [ 1 2 3 4 5 ],

The code can be used for characters and strings as well, by just replacing the datatype wherever required.

eg.

vector<char> a = {'d','g','y','u','t'};

To give

printing all 31 combinations
[ d ] , [ g ] , [ y ] , [ u ] , [ t ] , [ d g ] , [ d y ] , [ d u ] , [ d t ] , [ g y ] , [ g u ] , [ g t ] , [ y u ] , [ y t ] , 
[ u t ] , [ d g y ] , [ d g u ] , [ d g t ] , [ d y u ] , [ d y t ] , [ d u t ] , [ g y u ] , [ g y t ] , [ g u t ] , [ 
y u t ] , [ d g y u ] , [ d g y t ] , [ d g u t ] , [ d y u t ] , [ g y u t ] , [ d g y u t ] ,

and

vector<string> a = {"asdf","myfl", "itshot", "holy"};

to give

printing all 15 combinations
[ asdf ] , [ myfl ] , [ itshot ] , [ holy ] , [ asdf myfl ] , [ asdf itshot ] , [ asdf holy ] , [ myfl itshot ] , [ myfl holy ] , [ itshot holy ] , [ asdf myfl itshot ] , [ asdf myfl holy ] , [ asdf itshot holy ] , [ myfl itshot holy ] , [ asdf myfl itshot holy ] ,

• Mihail Malostanidis: They asked for permutations of N out of N (how many unique ways can they be rearranged?) (Usually, this ignores equality checking, assuming all values are "unique"). You provided indeed combinations of 1..=N out of N, preserving order. Two completely different words, two completely different problems.
• Aseem A. S.: I don't think the question was asked quite clearly, and that you read the question carefully. It mentions "I know that if the length of the array is n then there are n! possible combinations. How can this algorithm be written?" My guess is that the behold questioner wanted one, got two.

Answer 14

Permutation via iteration, in go.

// find all permutation of []T, return [][]T, not ordered, via iteration,
// tips: for empty input, return 1 empty combination,
func FindAllPermutation[T any](ts []T) [][]T {
 combList := [][]T{{}} // when empty input, has 1 empty combination, not 0 combination,
 for i := len(ts) - 1; i >= 0; i-- {
 // prefix := ts[0:i]
 mover := ts[i]
 // fmt.Printf("\nprefix = %v, mover = %v:\n", prefix, mover)
 var combList2 [][]T // combinations with an extra item added,
 for _, comb := range combList {
 for j := 0; j <= len(comb); j++ { // insert mover at index j of comb,
 comb2 := append(append(append([]T{}, comb[0:j]...), mover), comb[j:]...) // new_empty + left + mover + right
 // fmt.Printf("\t%v\n", comb2)
 combList2 = append(combList2, comb2)
 }
 }
 combList = combList2
 }
 return combList
}