generate_subset_bitmasking.md

Example Problem Statement:

Given an array nums = [1, 2, 3], the output should be all possible subsets:

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

Problem Understanding

The goal of the problem is to generate all possible subsets of a given list of numbers. Given an array nums, we need to find all possible combinations of its elements, including the empty subset and the subset containing all elements.

Intuition

Generating all subsets can be seen as a problem of generating all combinations. For a set with n elements, there are (2^n) possible subsets (including the empty subset). Each element in the array can either be included in a subset or not, which gives us a binary choice for each element.

Binary Representation of Subsets

A useful way to approach this problem is to use bit manipulation. Here's why:

• If we think of each subset as a combination of elements, each element can be either included or excluded.
• This inclusion/exclusion can be represented as a binary number where each bit represents whether an element is included (1) or not (0).

For example, for the array [1, 2, 3]:

• 000 (binary) represents the empty subset [].
• 001 (binary) represents the subset [1].
• 010 (binary) represents the subset [2].
• 011 (binary) represents the subset [1, 2].
• 100 (binary) represents the subset [3].
• 101 (binary) represents the subset [1, 3].
• 110 (binary) represents the subset [2, 3].
• 111 (binary) represents the subset [1, 2, 3].

Detailed Explanation of Code

Here's the code for generating all subsets:

def generate_subsets(nums):
    n = len(nums)
    subsets = []
    for i in range(1 << n):  # 2^n combinations
        subset = []
        for j in range(n):
            if i & (1 << j):  # Check if j-th bit is set
                subset.append(nums[j])
        subsets.append(subset)
    return subsets

nums = [1, 2, 3]
print(generate_subsets(nums))

Step-by-Step Breakdown

1. Initialize Variables:

n = len(nums)
subsets = []

• n is the length of the input array nums.
• subsets is an empty list that will store all the subsets.

2. Outer Loop: Iterate Over All Possible Combinations:

for i in range(1 << n):  # 2^n combinations

• 1 << n is equivalent to (2^n). It represents the total number of subsets.
• We iterate i from 0 to (2^n - 1). Each value of i represents a binary number that can be used to determine which elements are included in the subset.

3. Inner Loop: Determine Elements in the Current Subset:

subset = []
for j in range(n):
    if i & (1 << j):  # Check if j-th bit is set
        subset.append(nums[j])

• For each i, we initialize an empty list subset.
• We iterate over each bit position j from 0 to n-1.
• 1 << j creates a mask with only the j-th bit set.
• i & (1 << j) checks if the j-th bit of i is set (i.e., if the j-th element should be included in the subset).
• If the bit is set, we append nums[j] to the current subset.

4. Add the Current Subset to the List of Subsets:

subsets.append(subset)

5. Return the List of All Subsets:

return subsets

Example Walkthrough

Let's go through the example nums = [1, 2, 3]:

Initialization:

• n = 3
• subsets = []
• 1 << n = 8, so i will iterate from 0 to 7.

Iterations:

• i = 0 (binary 000): subset = [] (no bits set)
• i = 1 (binary 001): subset = [1] (only the 0th bit is set)
• i = 2 (binary 010): subset = [2] (only the 1st bit is set)
• i = 3 (binary 011): subset = [1, 2] (0th and 1st bits are set)
• i = 4 (binary 100): subset = [3] (only the 2nd bit is set)
• i = 5 (binary 101): subset = [1, 3] (0th and 2nd bits are set)
• i = 6 (binary 110): subset = [2, 3] (1st and 2nd bits are set)
• i = 7 (binary 111): subset = [1, 2, 3] (all bits are set)

At the end, subsets will contain all possible subsets of nums:

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

Complexity Analysis

• Time Complexity: (O(n * 2^n))
  • There are (2^n) subsets.
  • For each subset, we iterate through each element, leading to n iterations per subset.
• Space Complexity: (O(n * 2^n))
  • We store all (2^n) subsets, and each subset can have up to n elements.

This approach is efficient and leverages the power of bitwise operations to systematically generate all subsets of the input array.