Generating all possible Subsequences using Recursion

Difficulty Level : Easy
Last Updated : 19 Oct, 2020

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++

// 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 

Python3

# 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, [])  
  
#This code is contributed by Mayank Tyagi

Output:

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

Time Complexity:

$O(2^n)$

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