# 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 sequences
• We will run a loop from 0 to , 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 sequences in this way.

Below is the implementation of the approach:

## C++

 `// C++ implementation` `#include ` `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

 `// 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

 `# 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#

 `// 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`

## Javascript

 ``

Output:

```11
21
31
12
22
32
13
23
33```

Time Complexity: O(L*lenL), 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.

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