Subsets Generation Using Bitmask in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the outer loop for bitmask subsets generation.

DSA Javascript

for (let mask = 0; mask [1] (1 << nums.length); mask++) {
  // generate subset
}

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

A<=

B<

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= instead of < causes one extra iteration.

Using > or >= causes the loop to never run.

The loop runs from 0 to 2^n - 1, so mask must be less than 1 << nums.length.

2fill in blank

medium

Complete the inner loop to check if the j-th bit is set in the mask.

DSA Javascript

for (let j = 0; j < nums.length; j++) {
  if (mask & ([1])) {
    subset.push(nums[j]);
  }
}

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

A1 << j

Bj << 1

C1 >> j

Dj >> 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using j << 1 shifts j instead of 1, which is incorrect.

Using right shift operators does not check the correct bit.

To check if the j-th bit is set, use (1 << j) and bitwise AND with mask.

3fill in blank

hard

Fix the error in the code to correctly generate all subsets using bitmask.

DSA Javascript

function subsets(nums) {
  const result = [];
  for (let mask = 0; mask [1] (1 << nums.length); mask++) {
    const subset = [];
    for (let j = 0; j < nums.length; j++) {
      if (mask & (1 << j)) {
        subset.push(nums[j]);
      }
    }
    result.push(subset);
  }
  return result;
}

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

A<

B>

C<=

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= causes an extra empty subset at the end.

Using > or >= causes the loop to skip all iterations.

The outer loop must run while mask is less than 2^n, so use < not <=.

4fill in blank

hard

Fill both blanks to create a function that returns subsets of nums using bitmask.

DSA Javascript

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

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

A<

B<=

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= causes loops to run one extra time leading to errors.

Using > or >= causes loops to skip iterations.

Both loops should use < to iterate correctly over the range without extra iterations.

5fill in blank

hard

Fill all three blanks to generate subsets and print them.

DSA Javascript

function subsets(nums) {
  const result = [];
  for (let mask = 0; mask [1] (1 << nums.length); mask++) {
    const subset = [];
    for (let j = 0; j [2] nums.length; j++) {
      if (mask & ([3])) {
        subset.push(nums[j]);
      }
    }
    result.push(subset);
  }
  return result;
}

const nums = [1, 2, 3];
console.log(subsets(nums));

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

A<

B<=

C1 << j

Dj << 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= causes off-by-one errors.

Using j << 1 does not correctly check the bit.

Use < for loops and (1 << j) to check the j-th bit in mask.

Practice

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

easy

A. Only subsets with even number of elements

B. Only the largest subset

C. All possible subsets of the set

D. Only subsets with odd number of elements

Solution

1. Step 1: Understand bitmask representation

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

2. Step 2: Generate subsets using bitmask

   By iterating through all bitmasks from 0 to 2^n - 1, we get all possible combinations of elements, i.e., all subsets.

3. Final Answer:

   All possible subsets of the set -> Option C

4. Quick Check:

   Bitmask = subsets [OK]

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

Common Mistakes:

• Thinking bitmask generates only largest subsets
• Confusing subsets with permutations
• Assuming bitmask filters subsets by size

2. Which of the following is the correct way to start a loop for generating subsets using bitmask for an array of length n in JavaScript?

easy

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

B. for (let mask = 0; mask < (1 << n); mask++)

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

D. for (let mask = 1; 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 JavaScript.

2. Step 2: Correct loop syntax

   The loop should start at 0 and run while mask is less than (1 << n), incrementing mask by 1 each time.

3. Final Answer:

   for (let mask = 0; mask < (1 << n); mask++) -> Option B

4. Quick Check:

   Loop from 0 to 2^n-1 [OK]

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

Common Mistakes:

• Starting loop at 1 instead of 0
• Using mask <= n instead of mask < 2^n
• Confusing n with 2^n in loop condition

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

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

medium

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

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

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

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

Solution

1. Step 1: Understand bitmask iterations

   For arr length 2, mask runs from 0 to 3 (0b00 to 0b11), generating 4 subsets.

2. Step 2: Map mask bits to elements

   mask=0 (0b00) -> [], mask=1 (0b01) -> ["a"], mask=2 (0b10) -> ["b"], mask=3 (0b11) -> ["a", "b"]

3. Final Answer:

   [[], ["a"], ["b"], ["a", "b"]] -> Option A

4. Quick Check:

   All subsets of ["a", "b"] listed [OK]

Hint: Check bit positions for element inclusion [OK]

Common Mistakes:

• Missing empty subset []
• Swapping order of subsets
• Ignoring bitmask 0 producing empty subset

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

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

medium

A. The subsets array should be cleared inside the outer loop.

B. The inner loop should run until i <= arr.length.

C. The bitwise AND condition should be (mask | (1 << i)) instead of &.

D. The outer loop condition should be mask < (1 << arr.length), not <=.

Solution

1. Step 1: Check outer loop range

   The outer loop runs from 0 to (1 << arr.length) inclusive, which is one too many iterations (should be less than).

2. Step 2: Correct loop condition

   It should be mask < (1 << arr.length) to avoid an extra invalid subset iteration.

3. Final Answer:

   The outer loop condition should be mask < (1 << arr.length), not <=. -> Option D

4. Quick Check:

   Loop runs 2^n times, not 2^n+1 [OK]

Hint: Use <, not <=, for bitmask loop end [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Confusing bitwise AND (&) with OR (|)
• Extending inner loop beyond array length

5. Given an array arr = [1, 2, 3, 4], how many subsets will be generated by the bitmask method, and what is the subset generated when mask = 9? (mask is in decimal)

hard

A. 16 subsets; subset for mask 9 is [1, 4]

B. 8 subsets; subset for mask 9 is [2, 3]

C. 16 subsets; subset for mask 9 is [2, 4]

D. 15 subsets; subset for mask 9 is [1, 3, 4]

Solution

1. Step 1: Calculate total subsets for array length 4

   Total subsets = 2^4 = 16.

2. Step 2: Decode mask 9 in binary and map to elements

   9 in binary is 1001. Bits set at positions 0 and 3 correspond to arr[0] = 1 and arr[3] = 4.

3. Final Answer:

   16 subsets; subset for mask 9 is [1, 4] -> Option A

4. Quick Check:

   2^4 = 16 subsets; mask 9 = 1001 binary [OK]

Hint: Convert mask to binary, map bits to elements [OK]

Common Mistakes:

• Miscounting total subsets as 15
• Misinterpreting binary mask bits
• Confusing element indices with bit positions