Heap’s Algorithm for generating permutations

Heap’s algorithm is used to generate all permutations of n objects. The idea is to generate each permutation from the previous permutation by choosing a pair of elements to interchange, without disturbing the other n-2 elements. 
Following is the illustration of generating all the permutations of n given numbers.
Example:

Input: 1 2 3
Output: 1 2 3
        2 1 3
        3 1 2
        1 3 2
        2 3 1
        3 2 1

Algorithm: 

1. The algorithm generates (n-1)! permutations of the first n-1 elements, adjoining the last element to each of these. This will generate all of the permutations that end with the last element.
2. If n is odd, swap the first and last element and if n is even, then swap the i^th element (i is the counter starting from 0) and the last element and repeat the above algorithm till i is less than n.
3. In each iteration, the algorithm will produce all the permutations that end with the current last element.

Implementation: 

C++

// C++ program to print all permutations using
// Heap's algorithm
#include <bits/stdc++.h>
using namespace std;

// Prints the array
void printArr(int a[], int n)
{
    for (int i = 0; i < n; i++)
        cout << a[i] << " ";
    printf("\n");
}

// Generating permutation using Heap Algorithm
void heapPermutation(int a[], int size, int n)
{
    // if size becomes 1 then prints the obtained
    // permutation
    if (size == 1) {
        printArr(a, n);
        return;
    }

    for (int i = 0; i < size; i++) {
        heapPermutation(a, size - 1, n);

        // if size is odd, swap 0th i.e (first) and
        // (size-1)th i.e (last) element
        if (size % 2 == 1)
            swap(a[0], a[size - 1]);

        // If size is even, swap ith and
        // (size-1)th i.e (last) element
        else
            swap(a[i], a[size - 1]);
    }
}

// Driver code
int main()
{
    int a[] = { 1, 2, 3 };
    int n = sizeof a / sizeof a[0];
    heapPermutation(a, n, n);
    return 0;
}

Java

// Java program to print all permutations using
// Heap's algorithm
import java.io.*;
class HeapAlgo {
    // Prints the array
    void printArr(int a[], int n)
    {
        for (int i = 0; i < n; i++)
            System.out.print(a[i] + " ");
        System.out.println();
    }

    // Generating permutation using Heap Algorithm
    void heapPermutation(int a[], int size, int n)
    {
        // if size becomes 1 then prints the obtained
        // permutation
        if (size == 1)
            printArr(a, n);

        for (int i = 0; i < size; i++) {
            heapPermutation(a, size - 1, n);

            // if size is odd, swap 0th i.e (first) and
            // (size-1)th i.e (last) element
            if (size % 2 == 1) {
                int temp = a[0];
                a[0] = a[size - 1];
                a[size - 1] = temp;
            }

            // If size is even, swap ith
            // and (size-1)th i.e last element
            else {
                int temp = a[i];
                a[i] = a[size - 1];
                a[size - 1] = temp;
            }
        }
    }

    // Driver code
    public static void main(String args[])
    {
        HeapAlgo obj = new HeapAlgo();
        int a[] = { 1, 2, 3 };
        obj.heapPermutation(a, a.length, a.length);
    }
}

// This code has been contributed by Amit Khandelwal.

Python3

# Python program to print all permutations using
# Heap's algorithm

# Generating permutation using Heap Algorithm
def heapPermutation(a, size):

    # if size becomes 1 then prints the obtained
    # permutation
    if size == 1:
        print(a)
        return

    for i in range(size):
        heapPermutation(a, size-1)

        # if size is odd, swap 0th i.e (first)
        # and (size-1)th i.e (last) element
        # else If size is even, swap ith
        # and (size-1)th i.e (last) element
        if size & 1:
            a[0], a[size-1] = a[size-1], a[0]
        else:
            a[i], a[size-1] = a[size-1], a[i]


# Driver code
a = [1, 2, 3]
n = len(a)
heapPermutation(a, n)

# This code is contributed by ankush_953
# This code was cleaned up to by more pythonic by glubs9

C#

// C# program to print all permutations using
// Heap's algorithm
using System;

public class GFG {
    // Prints the array
    static void printArr(int[] a, int n)
    {
        for (int i = 0; i < n; i++)
            Console.Write(a[i] + " ");
        Console.WriteLine();
    }

    // Generating permutation using Heap Algorithm
    static void heapPermutation(int[] a, int size, int n)
    {
        // if size becomes 1 then prints the obtained
        // permutation
        if (size == 1)
            printArr(a, n);

        for (int i = 0; i < size; i++) {
            heapPermutation(a, size - 1, n);

            // if size is odd, swap 0th i.e (first) and
            // (size-1)th i.e (last) element
            if (size % 2 == 1) {
                int temp = a[0];
                a[0] = a[size - 1];
                a[size - 1] = temp;
            }

            // If size is even, swap ith and
            // (size-1)th i.e (last) element
            else {
                int temp = a[i];
                a[i] = a[size - 1];
                a[size - 1] = temp;
            }
        }
    }

    // Driver code
    public static void Main()
    {

        int[] a = { 1, 2, 3 };
        heapPermutation(a, a.Length, a.Length);
    }
}

/* This Java code is contributed by 29AjayKumar*/

Javascript

<script>

// JavaScript program to print all permutations using
// Heap's algorithm

// Prints the array
function printArr(a,n)
{
    document.write(a.join(" ")+"<br>");

}

// Generating permutation using Heap Algorithm
function heapPermutation(a,size,n)
{
        // if size becomes 1 then prints the obtained
        // permutation
        if (size == 1)
            printArr(a, n);

        for (let i = 0; i < size; i++) {
            heapPermutation(a, size - 1, n);

            // if size is odd, swap 0th i.e (first) and
            // (size-1)th i.e (last) element
            if (size % 2 == 1) {
                let temp = a[0];
                a[0] = a[size - 1];
                a[size - 1] = temp;
            }

            // If size is even, swap ith
            // and (size-1)th i.e last element
            else {
                let temp = a[i];
                a[i] = a[size - 1];
                a[size - 1] = temp;
            }
        }
}

// Driver code
let a=[1, 2, 3];
heapPermutation(a, a.length, a.length);


// This code is contributed by rag2127

</script>

Output

1 2 3 
2 1 3 
3 1 2 
1 3 2 
2 3 1 
3 2 1 

Time Complexity: O(N*N!), where N is the size of the given array.
Auxiliary Space: O(N), for recursive stack space of size N.

References:
1. “https://en.wikipedia.org/wiki/Heap%27s_algorithm#cite_note-3
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