Subsets Generation Using Bitmask in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the total number of subsets using bitmask.

DSA Typescript

const n = nums.length;
const totalSubsets = 1 [1] n;

Drag options to blanks, or click blank then click option'

A+

B<<

C-

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' or '*' instead of bit shift to calculate total subsets.

Using '-' which gives wrong result.

Using bit shift operator '<<' shifts 1 by n bits to get 2^n, which is total subsets count.

2fill in blank

medium

Complete the code to check if the j-th bit is set in the bitmask i.

DSA Typescript

if ((i & (1 [1] j)) !== 0) {
  subset.push(nums[j]);
}

Drag options to blanks, or click blank then click option'

A<<

B>>

C&

D|

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>>' which shifts bits right, not useful here.

Using '&' or '|' inside the shift operation incorrectly.

We use 1 << j to create a mask with only the j-th bit set, then & with i to check if that bit is set.

3fill in blank

hard

Fix the error in the code that generates subsets using bitmask.

DSA Typescript

for (let i = 0; i < [1]; i++) {
  const subset = [];
  for (let j = 0; j < n; j++) {
    if ((i & (1 << j)) !== 0) {
      subset.push(nums[j]);
    }
  }
  result.push(subset);
}

Drag options to blanks, or click blank then click option'

Aresult.length

Bn

Cnums.length

DtotalSubsets

Attempts:

3 left

💡 Hint

Common Mistakes

Using n or nums.length as loop limit, which is too small.

Using result.length which changes dynamically.

The outer loop should run from 0 to totalSubsets (2^n) to cover all subsets.

4fill in blank

hard

Fill both blanks to complete the function that returns all subsets using bitmask.

DSA Typescript

function subsets(nums: number[]): number[][] {
  const n = nums.length;
  const result: number[][] = [];
  const total = 1 [1] n;
  for (let i = 0; i < total; i++) {
    const subset: number[] = [];
    for (let j = 0; j < n; j++) {
      if ((i & (1 [2] j)) !== 0) {
        subset.push(nums[j]);
      }
    }
    result.push(subset);
  }
  return result;
}

Drag options to blanks, or click blank then click option'

A<<

B>>

D&

Attempts:

3 left

💡 Hint

Common Mistakes

Using right shift '>>' which is incorrect here.

Using bitwise AND '&' inside the shift operation.

Use '<<' to calculate total subsets and to shift 1 by j bits to check bit set.

5fill in blank

hard

Fill all three blanks to complete the code that generates subsets using bitmask with filtering subsets of size greater than 1.

DSA Typescript

function filteredSubsets(nums: number[]): number[][] {
  const n = nums.length;
  const result: number[][] = [];
  const total = 1 [1] n;
  for (let i = 0; i < total; i++) {
    const subset: number[] = [];
    for (let j = 0; j < n; j++) {
      if ((i & (1 [2] j)) !== 0) {
        subset.push(nums[j]);
      }
    }
    if (subset.length [3] 1) {
      result.push(subset);
    }
  }
  return result;
}

Drag options to blanks, or click blank then click option'

A<<

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>' for filtering subsets.

Using right shift '>>' which is incorrect here.

Use '<<' for bit shifts and '>' to filter subsets with length greater than 1.

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