# Subsets / Power-set Problem

In the Subsets problem, we find the power set of a given input using bitwise operations. In general, if we have n elements then the subsets are 2^n subsets.

Subsets problem, we need to write a program that finds all possible subsets (the power set) of a given input. The solution set must not contain duplicate subsets.

Subsets problem, we find the power set of given input without duplicates. Leetcode, one of the largest tech communities with hundreds of thousands of active users to participate and solve coding problems, listed this problem in their curriculum – the Leetcode Subsets Problem.

### Introduction

Let us discuss the subsets of a given input. This is one of the most popular questions asked in coding interviews.

Companies that have asked this in their coding interview are Apple, Microsoft, Amazon, Facebook, and many more.

### Problem Statement

We need to write a program that finds all possible input subsets ( the power set). The solution set must not contain duplicate subsets.

**Example 01:**

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

**Example 02:**

```
Input: [100]
Output: [[], [100]]
```

**Explanation:** The subsets of any given input are equal to its power set.

if, input n = 3, then, powerset => 2^n = 2^3 = 8.

Assume input has a length greater than or equal to 1.

**Hint: **Use the left-shift operator to achieve this.

### Thought Process

In this program, we find the power set of a given input using bitwise operations.

In general, if we have n elements then the subsets are 2^n subsets.

So for every possible case of having at least two elements, we can see that an element is present and not present in the subsets

Think of a solution that is iterative, uses bitwise operators, and generates the powerset.

Here is how we generate each subset using the outer-loop variable counter. Here is a table indicating how the value gets generated based on the counter input.

## HTML Table

Counter (in decimal) | Counter (in binary) | Subset |
---|---|---|

0 | 000 | [] |

1 | 001 | [1] |

2 | 010 | [2] |

3 | 011 | [1, 2] |

4 | 100 | [3] |

5 | 101 | [1, 3] |

6 | 110 | [2, 3] |

7 | 111 | [1, 2, 3] |

### Algorithm

We need to consider a counter variable that starts from 0 to 2^n - 1.

For every value, we are considering the binary representation and here we use the set bits in the binary representation to generate corresponding subsets.

- If all set bits are 0, then the corresponding subset is empty [].
- If the last bit is 1, then we put 1 in the subset as [1].

**Steps:**

We use two loops here, the outer-loop starts from 0 to 2^n - 1, and the inner loop continues to input array length n.

In the inner loop, we conditionally check (counter & (1 << j)) != 0), if yes, then we print the corresponding element from an array.

**Solution **Code

```
import java.util.ArrayList;
import java.util.List;
public class Subsets {
public static void main(String[] args) {
int[] input = {1, 2, 3};
System.out.println(subsets(input));
}
public static List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
int n = nums.length;
int powSize = (int) Math.pow(2, n);
for (int i = 0; i < powSize; i++) {
List<Integer> val = new ArrayList<>();
for (int j = 0; j < n; j++) {
if ((i & (1 << j)) != 0) {
val.add(nums[j]);
}
}
result.add(val);
}
return result;
}
}
```

### Complexity Analysis

**Time Complexity:**`O(n*2^n)`

, Time complexity is `n`

times the powerset.

**Space Complexity:**`O(2^n)`

, We are storing `2^`

subset elements in an array. So the extra space is directly proportional to *n*`O(2^`

.*n*)

### Where to go next?

If you want to learn more about bit manipulation, please check out the course Master Bit Manipulation for Coding Interviews