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 |
[3] [2] [2, 3] [1] [1, 3] [1, 2] [1, 2, 3]
Time Complexity:
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