Subsets Generation Using Bitmask in DSA Typescript - 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 list gets bigger.

Specifically, how does the number of steps change when we use bitmasking to create subsets?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function generateSubsets(nums: number[]): number[][] {
  const n = nums.length;
  const subsets: number[][] = [];
  for (let mask = 0; mask < (1 << n); mask++) {
    const subset: number[] = [];
    for (let i = 0; i < n; i++) {
      if ((mask & (1 << i)) !== 0) {
        subset.push(nums[i]);
      }
    }
    subsets.push(subset);
  }
  return subsets;
}
    

This code creates all possible subsets of the input array by using bitmasks to decide which elements to include.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The outer loop runs through all bitmasks from 0 to 2ⁿ - 1.
• Inner loop: For each bitmask, it checks each bit position from 0 to n-1.
• How many times: Outer loop runs 2ⁿ times, inner loop runs n times for each outer iteration.

How Execution Grows With Input

As the input size n grows, the number of subsets doubles for each new element added.

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

Pattern observation: The operations grow very fast, roughly doubling for each extra element, multiplied by the element count.

Final Time Complexity

Time Complexity: O(n * 2n)

This means the time needed grows very fast as the input size increases, because we must check all subsets and their elements.

Common Mistake

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

[OK] Correct: Each subset can have up to n elements, and we spend time checking each element for every subset, so the total work is multiplied by n.

Interview Connect

Understanding this time complexity helps you explain why generating all subsets is expensive and shows you can analyze nested loops with exponential growth clearly.

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 generate when applied to a set of elements?

easy

A. Only the largest subset

B. Only subsets with odd number of elements

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

2. Step 2: Generate subsets using bitmask

   By iterating through all bitmasks from 0 to 2^n - 1, we cover all combinations of elements, thus all subsets.

3. Final Answer:

   All possible subsets of the set -> Option D

4. Quick Check:

   Bitmask = All subsets [OK]

Hint: Bitmask from 0 to 2^n-1 covers all subsets [OK]

Common Mistakes:

• Thinking bitmask generates only even or odd subsets
• Confusing subsets with permutations
• Assuming bitmask generates only largest subsets

2. Which of the following TypeScript code snippets correctly initializes the bitmask loop for generating subsets of an array arr of length n?

easy

A. for (let mask = 0; mask <= n; mask++) {}

B. for (let mask = 1; mask <= n; mask++) {}

C. for (let mask = 0; mask < (1 << n); mask++) {}

D. for (let mask = 0; mask < n; mask++) {}

Solution

1. Step 1: Understand bitmask range

   Bitmask must cover all numbers from 0 to 2^n - 1, which is (1 << n) in code.

2. Step 2: Check loop correctness

   Loop from 0 to less than (1 << n) correctly iterates all bitmasks representing subsets.

3. Final Answer:

   for (let mask = 0; mask < (1 << n); mask++) {} -> Option C

4. Quick Check:

   Loop 0 to 2^n-1 = for (let mask = 0; mask < (1 << n); mask++) {} [OK]

Hint: Use (1 << n) for total subsets count [OK]

Common Mistakes:

• Using mask <= n instead of mask < 2^n
• Starting loop from 1 misses empty subset
• Using mask < n only covers single elements

3. What is the output of the following TypeScript code?

const arr = ["a", "b", "c"];
const subsets: string[][] = [];
const n = arr.length;
for (let mask = 0; mask < (1 << n); mask++) {
  const subset: string[] = [];
  for (let i = 0; i < n; i++) {
    if ((mask & (1 << i)) !== 0) {
      subset.push(arr[i]);
    }
  }
  subsets.push(subset);
}
console.log(subsets);

medium

A. [["a"], ["b", "a"], ["c", "b"], ["a", "b", "c"]]

B. [[], ["a"], ["b"], ["a", "b"], ["c"], ["a", "c"], ["b", "c"], ["a", "b", "c"]]

C. [[], ["a", "b", "c"]]

D. [["a"], ["b"], ["c"]]

Solution

1. Step 1: Understand bitmask iteration

   Loop runs from 0 to 7 (2^3 - 1), generating all subsets of 3 elements.

2. Step 2: Trace subsets generated

   Each mask selects elements where bit is set, producing all subsets including empty and full set.

3. Final Answer:

   [[], ["a"], ["b"], ["a", "b"], ["c"], ["a", "c"], ["b", "c"], ["a", "b", "c"]] -> Option B

4. Quick Check:

   All subsets of 3 elements = [[], ["a"], ["b"], ["a", "b"], ["c"], ["a", "c"], ["b", "c"], ["a", "b", "c"]] [OK]

Hint: Bitmask 0 to 7 covers all subsets of 3 elements [OK]

Common Mistakes:

• Missing empty subset []
• Incorrect order or missing subsets
• Confusing subsets with permutations

4. Identify the error in this TypeScript code for generating subsets using bitmask:

const arr = [1, 2, 3];
const n = arr.length;
for (let mask = 0; mask <= (1 << n); mask++) {
  const subset = [];
  for (let i = 0; i < n; i++) {
    if (mask & (1 << i)) {
      subset.push(arr[i]);
    }
  }
  console.log(subset);
}

medium

A. The outer loop should run while mask < (1 << n), not <=

B. The inner loop should run from i = 1 to n

C. The bitwise AND condition should be == 0 instead of truthy check

D. The subset array should be declared outside the loops

Solution

1. Step 1: Check outer loop range

   Loop runs from 0 to (1 << n) inclusive, which is one too many iterations (should be less than).

2. Step 2: Understand consequences

   When mask equals (1 << n), it exceeds the maximum bitmask for subsets, causing an extra invalid iteration.

3. Final Answer:

   The outer loop should run while mask < (1 << n), not <= -> Option A

4. Quick Check:

   Loop limit off by one = The outer loop should run while mask < (1 << n), not <= [OK]

Hint: Use mask < 2^n, not <= 2^n [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Changing inner loop range incorrectly
• Misusing bitwise condition checks

5. Given an array arr = [1, 2, 2], which code modification correctly generates all unique subsets using bitmask technique?

hard

A. Use a Set to store stringified subsets to avoid duplicates

B. Sort the array and skip adding duplicates inside inner loop

C. Increase bitmask range to (1 << (n+1)) to cover duplicates

D. No modification needed; bitmask handles duplicates automatically

Solution

1. Step 1: Understand duplicates in subsets

   Bitmask generates subsets by position, so duplicates in array cause duplicate subsets.

2. Step 2: Avoid duplicates using a Set

   Storing subsets as strings in a Set removes duplicates by keeping only unique subsets.

3. Final Answer:

   Use a Set to store stringified subsets to avoid duplicates -> Option A

4. Quick Check:

   Set removes duplicate subsets [OK]

Hint: Use Set to filter duplicate subsets from duplicates in array [OK]

Common Mistakes:

• Assuming bitmask alone removes duplicates
• Increasing bitmask range unnecessarily
• Trying to skip duplicates without sorting or tracking