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# Print all subsets of a given Set or Array

• Difficulty Level : Medium
• Last Updated : 28 Dec, 2022

Given a set of positive integers, find all its subsets.

Examples:

```Input: array = {1, 2, 3}
Output: // this space denotes null element.
1
1 2
1 2 3
1 3
2
2 3
3
Explanation: These are all the subsets that
can be formed using the array.

Input: 1 2
Output:
1
2
1 2
Explanation: These are all the subsets that
can be formed using the array.```
Recommended Practice

The iterative solution is already discussed here: the iterative approach to find all subsets. This article aims to provide a backtracking approach.

Approach: The idea is simple, that if there are n number of elements inside an array, there are n number of choices for the first element in the array. Moving forward to the next recursion call, there will be n-1 number of choices (as we cannot insert the last element again now) for inserting the second element in the array.
Using the above idea forms a recursive solution to the problem.

Algorithm:

1. Create a recursive function that takes the following parameters, input array, the current index, the output array, or current subset, if all the subsets need to be stored then a vector of the array is needed if the subsets need to be printed only then this space can be ignored.
2. First, print the subset (output array) that has been sent to the function and then run a for loop starting from the ‘index’ to n-1 where n is the size of the input array. We use a loop to demonstrate that we have exactly n number of choices to choose from when adding the first element to our new array.
3. Inside the loop, we call the function for the next index after inserting that particular index and then in the next call, we again have (n-1) choices to choose from and so it goes.
4. Whenever a call is made for the last index of the array : in that function call, the loop is not run as the condition i<A.size() is not fulfilled and hence, we backtrack to the last recursion call and call the function for the next index after removing the element at that current index.
1. We finally return when the initial loop ends and we have received all the subsets possible. Implementation:

## C++

 `// CPP program to find all subsets by backtracking. ` `#include ` `using` `namespace` `std; ` ` `  `// For our vector subset, at every step we have A.size()-i-1 ` `// choices to include. A loop helps us to choose each element ` `// and then move to the indices further present in the array. ` `// At the end we backtrack to choose a different index. ` `void` `subsetsUtil(vector<``int``>& A, vector >& res, ` `                 ``vector<``int``>& subset, ``int` `index) ` `{ ` `    ``res.push_back(subset); ` `    ``// Loop to choose from different elements present ` `    ``// after the current index 'index' ` `    ``for` `(``int` `i = index; i < A.size(); i++) { ` ` `  `        ``// include the A[i] in subset. ` `        ``subset.push_back(A[i]); ` ` `  `        ``// move onto the next element. ` `        ``subsetsUtil(A, res, subset, i + 1); ` ` `  `        ``// exclude the A[i] from subset and triggers ` `        ``// backtracking. ` `        ``subset.pop_back(); ` `    ``} ` ` `  `    ``return``; ` `} ` ` `  `// below function returns the subsets of vector A. ` `vector > subsets(vector<``int``>& A) ` `{ ` `    ``vector<``int``> subset; ` `    ``vector > res; ` ` `  `    ``// keeps track of current element in vector A ` `    ``// and the number of elements present in the array subset ` `    ``int` `index = 0; ` `    ``subsetsUtil(A, res, subset, index); ` ` `  `    ``return` `res; ` `} ` ` `  `// Driver Code. ` `int` `main() ` `{ ` `    ``// find the subsets of below vector. ` `    ``vector<``int``> array = { 1, 2, 3 }; ` ` `  `    ``// res will store all subsets. ` `    ``// O(2 ^ (number of elements inside array)) ` `    ``// because total number of subsets possible ` `    ``// are O(2^N) ` `    ``vector > res = subsets(array); ` ` `  `    ``// Print result ` `    ``for` `(``int` `i = 0; i < res.size(); i++) { ` `        ``for` `(``int` `j = 0; j < res[i].size(); j++) ` `            ``cout << res[i][j] << ``" "``; ` `        ``cout << endl; ` `    ``} ` ` `  `    ``return` `0; ` `} `

## Java

 `/*package whatever // do not write package name here */` ` `  `import` `java.io.*; ` `import` `java.util.*; ` `class` `GFG { ` `    ``public` `static` `void` `    ``findSubsets(List > subset, ArrayList nums, ArrayList output, ``int` `index) ` `    ``{ ` `        ``// Base Condition ` `        ``if` `(index == nums.size()) { ` `            ``subset.add(output); ` `            ``return``; ` `        ``} ` ` `  `        ``// Not Including Value which is at Index ` `        ``findSubsets(subset, nums, ``new` `ArrayList<>(output), index + ``1``); ` ` `  `        ``// Including Value which is at Index ` `        ``output.add(nums.get(index)); ` `        ``findSubsets(subset, nums, ``new` `ArrayList<>(output), index + ``1``); ` `    ``} ` ` `  `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` ` `  `        ``// Main List for storing all subsets ` `        ``List > subset = ``new` `ArrayList<>(); ` ` `  `        ``// Input ArrayList ` `        ``ArrayList input = ``new` `ArrayList<>(); ` `        ``input.add(``1``); ` `        ``input.add(``2``); ` `        ``input.add(``3``); ` ` `  `        ``findSubsets(subset, input, ``new` `ArrayList<>(), ``0``); ` ` `  `        ``// Comparator is used so that all subset get ` `        ``// sorted in ascending order of values ` `        ``Collections.sort(subset, (o1, o2) -> { ` `            ``int` `n = Math.min(o1.size(), o2.size()); ` `            ``for` `(``int` `i = ``0``; i < n; i++) { ` `                ``if` `(o1.get(i) == o2.get(i)) { ` `                    ``continue``; ` `                ``} ` `                ``else` `{ ` `                    ``// sort based on the unequal elements value ` `                    ``return` `o1.get(i) - o2.get(i); ` `                ``} ` `            ``} ` `            ``// sort based on size ` `            ``return` `o1.size() - o2.size(); ` `        ``}); ` ` `  `        ``// Printing Subset ` `        ``for` `(``int` `i = ``0``; i < subset.size(); i++) { ` `            ``for` `(``int` `j = ``0``; j < subset.get(i).size(); j++) { ` `                ``System.out.print(subset.get(i).get(j) + ``" "``); ` `            ``} ` `            ``System.out.println(); ` `        ``} ` `    ``} ` `}`

## Python3

 `# Python3 program to find all subsets ` `# by backtracking.  ` ` `  `# In the array A at every step we have two  ` `# choices for each element either we can  ` `# ignore the element or we can include the  ` `# element in our subset  ` `def` `subsetsUtil(A, subset, index): ` `    ``print``(``*``subset) ` `    ``for` `i ``in` `range``(index, ``len``(A)):  ` `         `  `        ``# include the A[i] in subset.  ` `        ``subset.append(A[i]) ` `         `  `        ``# move onto the next element.  ` `        ``subsetsUtil(A, subset, i ``+` `1``)  ` `         `  `        ``# exclude the A[i] from subset and  ` `        ``# triggers backtracking. ` `        ``subset.pop(``-``1``)  ` `    ``return` ` `  `# below function returns the subsets of vector A.  ` `def` `subsets(A): ` `    ``global` `res ` `    ``subset ``=` `[] ` `     `  `    ``# keeps track of current element in vector A  ` `    ``index ``=` `0` `    ``subsetsUtil(A, subset, index)  ` `     `  `# Driver Code ` ` `  `# find the subsets of below vector.  ` `array ``=` `[``1``, ``2``, ``3``] ` ` `  `# res will store all subsets.  ` `# O(2 ^ (number of elements inside array))  ` `# because at every step we have two choices  ` `# either include or ignore.  ` `subsets(array)  ` ` `  `# This code is contributed by SHUBHAMSINGH8410 `

## C#

 `/*package whatever // do not write package name here */` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `public` `class` `GFG { ` `    ``public` `static` `void` `    ``findSubsets(List > subset, List<``int``> nums, List<``int``> output, ``int` `index) ` `    ``{ ` `        ``// Base Condition ` `        ``if` `(index == nums.Count) { ` `            ``subset.Add(output); ` `            ``return``; ` `        ``} ` ` `  `        ``// Not Including Value which is at Index ` `        ``findSubsets(subset, nums, ``new` `List<``int``>(output), index + 1); ` ` `  `        ``// Including Value which is at Index ` `        ``output.Add(nums[index]); ` `        ``findSubsets(subset, nums, ``new` `List<``int``>(output), index + 1); ` `    ``} ` ` `  `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` ` `  `        ``// Main List for storing all subsets ` `        ``List > subset = ``new` `List >(); ` ` `  `        ``// Input List ` `        ``List<``int``> input = ``new` `List<``int``>(); ` `        ``input.Add(1); ` `        ``input.Add(2); ` `        ``input.Add(3); ` ` `  `        ``findSubsets(subset, input, ``new` `List<``int``>(), 0); ` ` `  `        ``// Comparator is used so that all subset get ` `        ``// sorted in ascending order of values ` `        ``subset.Sort((o1, o2) = > { ` `            ``int` `n = Math.Min(o1.Count, o2.Count); ` `            ``for` `(``int` `i = 0; i < n; i++) { ` `                ``if` `(o1[i] == o2[i]) { ` `                    ``continue``; ` `                ``} ` `                ``else` `{ ` `                    ``return` `o1[i] - o2[i]; ` `                ``} ` `            ``} ` `            ``return` `1; ` `        ``}); ` ` `  `        ``// Printing Subset ` `        ``for` `(``int` `i = 0; i < subset.Count; i++) { ` `            ``for` `(``int` `j = 0; j < subset[i].Count; j++) { ` `                ``Console.Write(subset[i][j] + ``" "``); ` `            ``} ` `            ``Console.WriteLine(); ` `        ``} ` `    ``} ` `} ` ` `  `// This code is contributed by shikhasingrajput`

## Javascript

 ``

Output

```1
1 2
1 2 3
1 3
2
2 3
3 ```

Complexity Analysis:

• Time Complexity: O(n. 2^n). Total number of subsets generated are 2^n, So Time Complexity is O(2^n). If we include the time taken to copy the subset vector into the res vector the time taken will be equal to the size of the subset vector.
• Auxiliary Space: O(n) There can be at max n recursion calls at a particular time, which would consume O(n) stack space.

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