## Print out all permutations of an Array

print all permutations of a string c++

permutation of numbers in java

permutations of an array of arrays

number of permutations of a string

permutation of array python

how to get all permutations of an array python

python program to print all permutations of a string

**This question already has answers here**:

Creating (or printing) the permutations of an array is much easier done as a combination of recursively and iteratively than purely iteratively. There are surely iterative ways to do it, but it is particularly simple with a combination. Specifically, note that there are by definition N! permutations of a length N array - N choices for the first slot, N-1 choices for the 2nd, etc etc. So, we can break an algorithm down into two steps *for each index i in the array*.

- Select an element in the sub-array
`arr[i....end]`

to be the`ith`

element of the array. Swap that element with the element currently at`arr[i]`

. - Recursively permute
`arr[i+1...end]`

.

We note that this will run in O(N!), as on the 1st call N sub calls will be made, each of which will make N-1 sub calls, etc etc. Moreover, every element will end up being in every position, and so long as only swaps are made no element will ever be duplicated.

public static void permute(int[] arr){ permuteHelper(arr, 0); } private static void permuteHelper(int[] arr, int index){ if(index >= arr.length - 1){ //If we are at the last element - nothing left to permute //System.out.println(Arrays.toString(arr)); //Print the array System.out.print("["); for(int i = 0; i < arr.length - 1; i++){ System.out.print(arr[i] + ", "); } if(arr.length > 0) System.out.print(arr[arr.length - 1]); System.out.println("]"); return; } for(int i = index; i < arr.length; i++){ //For each index in the sub array arr[index...end] //Swap the elements at indices index and i int t = arr[index]; arr[index] = arr[i]; arr[i] = t; //Recurse on the sub array arr[index+1...end] permuteHelper(arr, index+1); //Swap the elements back t = arr[index]; arr[index] = arr[i]; arr[i] = t; } }

Sample input, output:

public static void main(String[] args) { permute(new int[]{1,2,3,4}); } [1, 2, 3, 4] [1, 2, 4, 3] [1, 3, 2, 4] [1, 3, 4, 2] [1, 4, 3, 2] [1, 4, 2, 3] [2, 1, 3, 4] [2, 1, 4, 3] [2, 3, 1, 4] [2, 3, 4, 1] [2, 4, 3, 1] [2, 4, 1, 3] [3, 2, 1, 4] [3, 2, 4, 1] [3, 1, 2, 4] [3, 1, 4, 2] [3, 4, 1, 2] [3, 4, 2, 1] [4, 2, 3, 1] [4, 2, 1, 3] [4, 3, 2, 1] [4, 3, 1, 2] [4, 1, 3, 2] [4, 1, 2, 3]

**Write a program to print all permutations of a given string ,** A quick and practical guide to generating array permutations in Java. of n elements has n! permutations. Here n! is the factorial, which is the product of all positive integers smaller or equal to n. System.out.print(input[i]);. }. Given an array arr[] of size N, the task is to generate and print all permutations of the given array. Examples: Input: arr[] = {1, 2} Output: 1 2 2 1

I have followed this method most of the time .. (it's given by the Robert Sedgewick and Kevin Wayne. ).

public class Permutations { // print N! permutation of the characters of the string s (in order) public static void perm1(String s) { perm1("", s); } private static void perm1(String prefix, String s) { int N = s.length(); if (N == 0) System.out.println(prefix); else { for (int i = 0; i < N; i++) perm1(prefix + s.charAt(i), s.substring(0, i) + s.substring(i+1, N)); } } // print N! permutation of the elements of array a (not in order) public static void perm2(String s) { int N = s.length(); char[] a = new char[N]; for (int i = 0; i < N; i++) a[i] = s.charAt(i); perm2(a, N); } private static void perm2(char[] a, int n) { if (n == 1) { System.out.println(a); return; } for (int i = 0; i < n; i++) { swap(a, i, n-1); perm2(a, n-1); swap(a, i, n-1); } } // swap the characters at indices i and j private static void swap(char[] a, int i, int j) { char c; c = a[i]; a[i] = a[j]; a[j] = c; }

However There is also an easier way to do this. May be you can work also around this

class PermutingArray { static void permutingArray(java.util.List<Integer> arrayList, int element) { for (int i = element; i < arrayList.size(); i++) { java.util.Collections.swap(arrayList, i, element); permutingArray(arrayList, element + 1); java.util.Collections.swap(arrayList, element, i); } if (element == arrayList.size() - 1) { System.out.println(java.util.Arrays.toString(arrayList.toArray())); } } public static void main(String[] args) { PermutingArray .permutingArray(java.util.Arrays.asList(9, 8, 7, 6, 4), 0); } }

Working Example here .. IDeone Link

**Permutations of an Array in Java,** Creating (or printing) the permutations of an array is much easier done as a combination of recursively and iteratively than purely iteratively. array = [1, 2, 3, 4] function permutation(start, end): if end==start: print array return for i -> (start, end+1): for i -> (start, end+1): permutation(start+1,end) Now, 1234 will be printed as it is the first permutation of the number 1, 2, 3 and 4. Till now, we are able to implement the logic of recursion.

The trick is to return a special value (`false`

in the code below) from `nextPerm`

when it was the last permutation (i.e. when array become sorted in descending order):

import java.util.*; public class Main { public static boolean nextPerm(List<Integer> a) { int i = a.size() - 2; while (i >= 0 && a.get(i) >= a.get(i + 1)) i--; if (i < 0) return false; int j = a.size() - 1; while (a.get(i) >= a.get(j)) j--; Collections.swap(a, i, j); Collections.reverse(a.subList(i + 1, a.size())); return true; } ...

Then you can use the loop (note that the array required be sorted in ascending order initially):

... public static void main(String[] args) { List<Integer> a = Arrays.asList(new Integer[] {1, 2, 3, 4}); do { System.out.println(a); } while (nextPerm(a)); } }

You can try this code here: http://ideone.com/URDFsc

**Print out all permutations of an Array,** This post is about printing all the permutations of an array with the use of recursion. This is also a very common question of computer Given array of distinct integers, print all permutations of the array. For example: array : [10, 20, 30] Permuations are : [10, 20, 30] [10, 30, 20] [20, 10, 30] [20, 30, 10] [30, 10, 20]

**Generating permutations of all elements of an array,** This tutorial provides how to print all permutations of array in java. Given array of distinct integers, print all permutations of the array. Print all the permutation of length L using the elements of an array | Iterative Print the sequence of size N in which every term is sum of previous K terms Count number of permutation of an Array having no SubArray of size two or more from original Array

**Permutations of array in java,** Given a collection of numbers, return all possible permutations. num.length; i++) { //list of list in current iteration of the array num ArrayList<ArrayList<Integer>> Print all distinct permutations of a given string with duplicates. Permutations of a given string using STL Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.

**LeetCode – Permutations (Java),** For example these are all the permutations of three elements: {1,2,3} {1,3,2} {2,1,3} public void permutation1(int[] a, int n) { if(n == 1) { System.out.println(Arrays. The algorithm above works but the output can be improved. There are many ways to generate all permutations of an array. In this article, we saw the recursive and iterative Heap's algorithm and how to generate a sorted list of permutations. It's not feasible to generate all permutations for large arrays, therefore, we can generate random permutations instead.

##### Comments

- Sample input: Enter Length of Array: 5
- Output: (1, 3, 5, 2, 4) (1, 4, 2, 5, 3) (2, 4, 1, 3, 5) (2, 5, 3, 1, 4) (3, 1, 4, 2, 5) (3, 5, 2, 4, 1) (4, 1, 3, 5, 2) (4, 2, 5, 3, 1) (5, 2, 4, 1, 3) (5, 3, 1, 4, 2)
- That is the hidden meaning of this piece of code:
`int[] B = new int[pivot+1]; reverseArray(B);`

? This reverses the zeroed array, doesn't it???`reverseArray(A)`

- this also looks strange... I suggest you to use the implementation from my answer instead, I tested it well, and it works as expected. - I'm trying to do this without using the package java.utl.Arrays
- It doesn't use util.arrays...
- My function can only have parameters Int[] A
- 1) Replaced Arrays.toString(..) with the by-hand version. It's literally the same thing, just doing it out. 2) The permute function
*does*only have parameter a single parameter of type`int[]`

. Is creating helper functions off limits? - This is an amazing answer! Thank you so much
- The second solution is so compact! Shows how good you are at Java API usage.