Subsets Generation Using Bitmask in DSA Python - Time & Space Complexity

Time Complexity: Subsets Generation Using Bitmask

O(n * 2^n)

Understanding Time Complexity

We want to understand how the time needed to generate all subsets grows as the input size increases.

How does the number of operations change when we create subsets using bitmasks?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

nums = [1, 2, 3]
subsets = []
n = len(nums)
for mask in range(1 << n):
    subset = []
    for i in range(n):
        if mask & (1 << i):
            subset.append(nums[i])
    subsets.append(subset)
    

This code generates all subsets of the list nums using bitmasking.

Identify Repeating Operations

• Primary operation: The nested loops: outer loop runs over all bitmasks, inner loop checks each bit.
• How many times: Outer loop runs 2ⁿ times, inner loop runs n times for each outer iteration.

How Execution Grows With Input

As input size grows, the number of subsets doubles for each new element, and each subset requires checking all elements.

Input Size (n)	Approx. Operations
10	About 10 * 210 = 10,240
20	About 20 * 220 = 20,971,520
30	About 30 * 230 = 32,212,254,720

Pattern observation: The operations grow very fast, doubling with each added element and multiplying by n.

Final Time Complexity

Time Complexity: O(n * 2^n)

This means the time needed grows very quickly as input size increases, because we check every subset and every element.

Common Mistake

[X] Wrong: "Generating subsets is just O(2^n) because there are 2^n subsets."

[OK] Correct: We also spend time checking each element for every subset, so the inner loop adds a factor of n, making it O(n * 2^n).

Interview Connect

Understanding this complexity helps you explain why generating all subsets is expensive and shows your grasp of nested loops and exponential growth.

Self-Check

"What if we only generated subsets of a fixed size k instead of all sizes? How would the time complexity change?"

Practice

1. What does the bitmask technique help to generate when applied to a set of elements?

easy

A. Only subsets with odd number of elements

B. Only the largest subset of the set

C. Only subsets with even number of elements

D. All possible subsets of the set

Solution

1. Step 1: Understand bitmask representation

   Each bit in a bitmask represents whether an element is included (1) or excluded (0) in a subset.

2. Step 2: Generate subsets using bitmask

   By iterating from 0 to 2^n - 1 (where n is number of elements), all combinations of included/excluded elements are covered, producing all subsets.

3. Final Answer:

   All possible subsets of the set -> Option D

4. Quick Check:

   Bitmask = subsets generator [OK]

Hint: Bitmask bits map directly to subset inclusion [OK]

Common Mistakes:

• Thinking bitmask generates only large subsets
• Confusing bitmask with sorting or filtering
• Assuming bitmask generates only even or odd subsets

2. Which of the following is the correct Python syntax to iterate over all bitmasks for a set of size n?

easy

A. for mask in range(1 << n):

B. for mask in range(n << 1):

C. for mask in range(2 * n):

D. for mask in range(n ** 2):

Solution

1. Step 1: Understand bit shift for 2^n

   Using 1 << n shifts 1 left by n bits, which equals 2 to the power n.

2. Step 2: Use range to cover all bitmasks

   Iterating from 0 to 1 << n - 1 covers all subsets represented by bitmasks.

3. Final Answer:

   for mask in range(1 << n): -> Option A

4. Quick Check:

   1 << n = 2^n range [OK]

Hint: Use 1 << n for 2^n subsets [OK]

Common Mistakes:

• Using multiplication instead of bit shift
• Using n squared or doubled incorrectly
• Confusing range limits

3. What is the output of the following code?

arr = ["a", "b"]
subsets = []
for mask in range(1 << len(arr)):
    subset = []
    for i in range(len(arr)):
        if mask & (1 << i):
            subset.append(arr[i])
    subsets.append(subset)
print(subsets)

medium

A. [['a', 'b'], ['a'], ['b'], []]

B. [[], ['a'], ['b'], ['a', 'b']]

C. [[], ['b'], ['a'], ['b', 'a']]

D. [['a'], ['b'], ['a', 'b'], []]

Solution

1. Step 1: Understand bitmask iteration

   For arr of length 2, masks go from 0 to 3 (00, 01, 10, 11 in binary).

2. Step 2: Map bits to elements

   Mask 0 (00) -> no elements -> [] Mask 1 (01) -> include arr[0] -> ['a'] Mask 2 (10) -> include arr[1] -> ['b'] Mask 3 (11) -> include arr[0], arr[1] -> ['a', 'b']

3. Final Answer:

   [[], ['a'], ['b'], ['a', 'b']] -> Option B

4. Quick Check:

   Bitmask subsets match output [OK]

Hint: Trace bitmask bits to elements included [OK]

Common Mistakes:

• Mixing order of subsets
• Appending elements incorrectly
• Confusing bit positions

4. Identify the error in the following code snippet for generating subsets using bitmask:

arr = [1, 2, 3]
subsets = []
for mask in range(1 << len(arr)):
    subset = []
    for i in range(len(arr)):
        if mask & i:
            subset.append(arr[i])
    subsets.append(subset)
print(subsets)

medium

A. The subset list should be initialized outside the loop

B. The range should be 'range(len(arr))' instead of 'range(1 << len(arr))'

C. The condition should be 'mask & (1 << i)' instead of 'mask & i'

D. The print statement should be inside the inner loop

Solution

1. Step 1: Analyze bitmask condition

   The condition 'mask & i' incorrectly checks bits; it should check if the i-th bit is set using 'mask & (1 << i)'.

2. Step 2: Correct bit checking

   Using 'mask & (1 << i)' correctly tests if the bit at position i is 1, indicating inclusion of arr[i].

3. Final Answer:

   The condition should be 'mask & (1 << i)' instead of 'mask & i' -> Option C

4. Quick Check:

   Bit check needs shift [OK]

Hint: Use bit shift to check bits, not index alone [OK]

Common Mistakes:

• Using 'mask & i' instead of 'mask & (1 << i)'
• Initializing subset outside loop causing accumulation
• Misplacing print inside loops

5. Given a list arr = [1, 2, 2], what is the output of the following code that generates subsets using bitmask?

arr = [1, 2, 2]
subsets = []
for mask in range(1 << len(arr)):
    subset = []
    for i in range(len(arr)):
        if mask & (1 << i):
            subset.append(arr[i])
    subsets.append(subset)
print(subsets)

hard

A. [[], [1], [2], [1, 2], [2], [1, 2], [2, 2], [1, 2, 2]]

B. [[], [1], [2], [2], [1, 2], [1, 2], [2, 2], [1, 2, 2]]

C. [[], [1], [2], [1, 2], [2, 2], [1, 2, 2]]

D. [[], [1], [2], [1, 2], [2, 2], [1, 2, 2], [2]]

Solution

1. Step 1: Calculate total subsets

   For 3 elements, total subsets = 2^3 = 8, masks from 0 to 7.

2. Step 2: Map each mask to subset

   Mask 0 (000): [] Mask 1 (001): [1] Mask 2 (010): [2] (second element) Mask 3 (011): [1, 2] Mask 4 (100): [2] (third element) Mask 5 (101): [1, 2] Mask 6 (110): [2, 2] Mask 7 (111): [1, 2, 2]

3. Final Answer:

   [[], [1], [2], [1, 2], [2], [1, 2], [2, 2], [1, 2, 2]] -> Option A

4. Quick Check:

   Duplicates appear as separate elements [OK]

Hint: Duplicates create repeated subsets with same elements [OK]

Common Mistakes:

• Assuming duplicates merge subsets
• Missing subsets with repeated elements
• Incorrect mask to subset mapping