Generating all possible Subsequences using Recursion

Given an array. The task is to generate and print all of the possible subsequences of the given array using recursion.

Examples:

Input : [1, 2, 3]
Output : [3], [2], [2, 3], [1], [1, 3], [1, 2], [1, 2, 3]

Input : [1, 2]
Output : [2], [1], [1, 2]

Approach: For every element in the array, there are two choices, either to include it in the subsequence or not include it. Apply this for every element in the array starting from index 0 until we reach the last index. Print the subsequence once the last index is reached.



Below diagram shows the recursion tree for array, arr[] = {1, 2}.

Below is the implementation of the above approach.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ code to print all possible
// subsequences for given array using
// recursion
#include <bits/stdc++.h>
using namespace std;
  
void printArray(vector<int> arr, int n)
{
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
    cout << endl;
}
  
// Recursive function to print all
// possible subsequences for given array
void printSubsequences(vector<int> arr, int index, 
                       vector<int> subarr)
{
    // Print the subsequence when reach
    // the leaf of recursion tree
    if (index == arr.size())
    {
        int l = subarr.size();
  
        // Condition to avoid printing
        // empty subsequence
        if (l != 0)
            printArray(subarr, l);
    }
    else
    {
        // Subsequence without including
        // the element at current index
        printSubsequences(arr, index + 1, subarr);
  
        subarr.push_back(arr[index]);
  
        // Subsequence including the element
        // at current index
        printSubsequences(arr, index + 1, subarr);
    }
    return;
}
  
// Driver Code
int main()
{
    vector<int> arr{1, 2, 3};
    vector<int> b;
  
    printSubsequences(arr, 0, b);
  
    return 0;
}
  
// This code is contributed by
// sanjeev2552

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 code to print all possible 
# subsequences for given array using 
# recursion
  
# Recursive function to print all
# possible subsequences for given array
def printSubsequences(arr, index, subarr):
      
    # Print the subsequence when reach 
    # the leaf of recursion tree
    if index == len(arr):
          
        # Condition to avoid printing
        # empty subsequence
        if len(subarr) != 0:
            print(subarr)
      
    else:
        # Subsequence without including 
        # the element at current index
        printSubsequences(arr, index + 1, subarr)
          
        # Subsequence including the element
        # at current index
        printSubsequences(arr, index + 1, \
                            subarr+[arr[index]])
      
    return
          
arr = [1, 2, 3]
  
printSubsequences(arr, 0, [])

chevron_right


Output:

[3]
[2]
[2, 3]
[1]
[1, 3]
[1, 2]
[1, 2, 3]

Time Complexity: O(2^n)



My Personal Notes arrow_drop_up

I like solving puzzles

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : sanjeev2552