Print all the permutation of length L using the elements of an array | Iterative

Given an array of unique elements, we have to find all the permutations of length L using the elements of the array. Repetition of elements is allowed.
Examples: 

Input: arr = { 1, 2 }, L=3 
Output: 
111 
211 
121 
221 
112 
212 
122 
222
Input: arr = { 1, 2, 3 }, L=2 
Output: 
11 
21 
31 
12 
22 
32 
13 
23 
33 
 

Approach: 

• To form a sequence of length L with N number of elements, it is known that the i-th element of the sequence can be filled in N ways. So there will be N^{L}   sequences
• We will run a loop from 0 to (N^{L} - 1)   , for every i we will convert i from base 10 to base N. The digits of the converted number will represent the indices of the array
• We can print all the N^{L}   sequences in this way.

Below is the implementation of the approach: 

// C++ implementation
#include <bits/stdc++.h>
using namespace std;

// Convert the number to Lth
// base and print the sequence
void convert_To_Len_th_base(int n,
                            int arr[],
                            int len,
                            int L)
{
    // Sequence is of length L
    for (int i = 0; i < L; i++) {
        // Print the ith element
        // of sequence
        cout << arr[n % len];
        n /= len;
    }
    cout << endl;
}

// Print all the permuataions
void print(int arr[],
           int len,
           int L)
{
    // There can be (len)^l
    // permutations
    for (int i = 0; i < (int)pow(len, L); i++) {
        // Convert i to len th base
        convert_To_Len_th_base(i, arr, len, L);
    }
}

// Driver code
int main()
{
    int arr[] = { 1, 2, 3 };
    int len = sizeof(arr) / sizeof(arr[0]);
    int L = 2;

    // function call
    print(arr, len, L);

    return 0;
}

// Java implementation for above approach
import java.io.*;

class GFG 
{
    
// Convert the number to Lth
// base and print the sequence
static void convert_To_Len_th_base(int n, int arr[], 
                                   int len, int L)
{
    // Sequence is of length L
    for (int i = 0; i < L; i++) 
    {
        // Print the ith element
        // of sequence
        System.out.print(arr[n % len]);
        n /= len;
    }
    System.out.println();
}

// Print all the permuataions
static void print(int arr[], int len, int L)
{
    // There can be (len)^l
    // permutations
    for (int i = 0; 
             i < (int)Math.pow(len, L); i++) 
    {
        // Convert i to len th base
        convert_To_Len_th_base(i, arr, len, L);
    }
}

// Driver code
public static void main (String[] args) 
{
    int arr[] = { 1, 2, 3 };
    int len = arr.length;
    int L = 2;
    
    // function call
    print(arr, len, L);
}
}

// This code is contributed by ajit. 

# Python3 implementation for the above approach

# Convert the number to Lth
# base and print the sequence
def convert_To_Len_th_base(n, arr, Len, L):
    
    # Sequence is of Length L
    for i in range(L):
        
        # Print the ith element
        # of sequence
        print(arr[n % Len], end = "")
        n //= Len
    print()

# Print all the permuataions
def printf(arr, Len, L):
    
    # There can be (Len)^l permutations
    for i in range(pow(Len, L)):
        
        # Convert i to Len th base
        convert_To_Len_th_base(i, arr, Len, L)

# Driver code
arr = [1, 2, 3]
Len = len(arr)
L = 2

# function call
printf(arr, Len, L)

# This code is contributed by Mohit Kumar

// C# implementation for above approach
using System;

class GFG 
{
    
// Convert the number to Lth
// base and print the sequence
static void convert_To_Len_th_base(int n, int []arr, 
                                   int len, int L)
{
    // Sequence is of length L
    for (int i = 0; i < L; i++) 
    {
        // Print the ith element
        // of sequence
        Console.Write(arr[n % len]);
        n /= len;
    }
    Console.WriteLine();
}

// Print all the permuataions
static void print(int []arr, int len, int L)
{
    // There can be (len)^l
    // permutations
    for (int i = 0; 
            i < (int)Math.Pow(len, L); i++) 
    {
        // Convert i to len th base
        convert_To_Len_th_base(i, arr, len, L);
    }
}

// Driver code
public static void Main (String[] args) 
{
    int []arr = { 1, 2, 3 };
    int len = arr.Length;
    int L = 2;
    
    // function call
    print(arr, len, L);
}
}

// This code is contributed by Rajput-Ji

<script>
// Javascript implementation

// Convert the number to Lth
// base and print the sequence
function convert_To_Len_th_base( n, arr, len, L)
{
    // Sequence is of length L
    for ( i = 0; i < L; i++) {
        // Print the ith element
        // of sequence
        document.write(parseInt(arr[n % len]));
        n = parseInt(n / len);
    }
   document.write("<br>");
}

// Print all the permuataions
function print( arr, len, L)
{
    // There can be (len)^l
    // permutations
    for (var i = 0; i < parseInt(Math.pow(len, L)); i++) {
        // Convert i to len th base
        convert_To_Len_th_base(i, arr, len, L);
    }
}

var arr = [ 1, 2, 3 ];
var len = arr.length;
var L = 2;

// function call
print(arr, len, L);

//This code is contributed by SoumikModnal
</script>

11
21
31
12
22
32
13
23
33

Time Complexity: O(L*len^L), as we are using a loop to traverse pow(len,L) times and in each traversal, we are calling the function convert_To_Len_th_base which will cost O(L). Where L is the number of elements in the array.
Auxiliary Space: O(1), as we are not using any extra space.

andrew1234

Follow

Improve

Article Tags :

• Mathematical
• Combinatorial
• Competitive Programming
• DSA
• permutation

+1 More

Practice Tags :

• Combinatorial
• Mathematical
• permutation