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;// Recursive function to print all// possible subsequences for given arrayvoid printSubsequences(int arr[], int index, vector<int> &subarr,int n){ // Print the subsequence when reach // the leaf of recursion tree if (index == n) { for (auto it:subarr){ cout << it << " "; } if(subarr.size()==0) cout<<"{}"; cout<<endl; return; } else { //pick the current index into the subsequence. subarr.push_back(arr[index]); printSubsequences(arr, index + 1, subarr,n); subarr.pop_back(); //not picking the element into the subsequence. printSubsequences(arr, index + 1, subarr,n); } }// Driver Codeint main(){ int arr[]={1, 2, 3}; int n=sizeof(arr)/sizeof(arr[0]); vector<int> vec; printSubsequences(arr, 0, vec,n); return 0;}// This code is contributed by// vivekr4400 |
Java
// Java code to print all possible// subsequences for given array using// recursionimport java.io.*;import java.util.*;class GFG{ // Recursive function to print all// possible subsequences for given arraypublic static void printSubsequences(int[] arr, int index, ArrayList<Integer> path){ // Print the subsequence when reach // the leaf of recursion tree if (index == arr.length) { // Condition to avoid printing // empty subsequence if (path.size() > 0) System.out.println(path); } else { // Subsequence without including // the element at current index printSubsequences(arr, index + 1, path); path.add(arr[index]); // Subsequence including the element // at current index printSubsequences(arr, index + 1, path); // Backtrack to remove the recently // inserted element path.remove(path.size() - 1); } return;}// Driver codepublic static void main(String[] args){ int[] arr = { 1, 2, 3 }; // Auxiliary space to store each path ArrayList<Integer> path = new ArrayList<>(); printSubsequences(arr, 0, path);}}// This code is contributed by Mukul Sharma |
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 |
C#
// C# code to print all possible// subsequences for given array using// recursionusing System;using System.Collections.Generic;class GFG { // Recursive function to print all // possible subsequences for given array static void printSubsequences(int[] arr, int index, List<int> path) { // Print the subsequence when reach // the leaf of recursion tree if (index == arr.Length) { // Condition to avoid printing // empty subsequence if (path.Count > 0) { Console.Write("["); for(int i = 0; i < path.Count - 1; i++) { Console.Write(path[i] + ", "); } Console.WriteLine(path[path.Count - 1] + "]"); } } else { // Subsequence without including // the element at current index printSubsequences(arr, index + 1, path); path.Add(arr[index]); // Subsequence including the element // at current index printSubsequences(arr, index + 1, path); // Backtrack to remove the recently // inserted element path.RemoveAt(path.Count - 1); } return; } static void Main() { int[] arr = { 1, 2, 3 }; // Auxiliary space to store each path List<int> path = new List<int>(); printSubsequences(arr, 0, path); }}// This code is contributed by rameshtravel07. |
Javascript
<script>// Javascript code to print all possible// subsequences for given array using// recursion// Recursive function to print all// possible subsequences for given arrayfunction printSubsequences(arr, index, path){ // Print the subsequence when reach // the leaf of recursion tree if (index == arr.length) { // Condition to avoid printing // empty subsequence if (path.length > 0) document.write(`[${path}]<br>`); } else { // Subsequence without including // the element at current index printSubsequences(arr, index + 1, path); path.push(arr[index]); // Subsequence including the element // at current index printSubsequences(arr, index + 1, path); // Backtrack to remove the recently // inserted element path.pop(); } return;}// Driver codelet arr = [1, 2, 3];// Auxiliary space to store each pathlet path = new Array();printSubsequences(arr, 0, path);// This code is contributed by gfgking</script> |
output
1 2 3
1 2
1 3
1
2 3
2
3
{}
Time Complexity: 
Space Complexity:
O(n) , Because of the recursion stack.
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