Subsets Generation Using Bitmask in DSA Javascript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind generating subsets using bitmask?

Each subset corresponds to a binary number where each bit represents whether an element is included (1) or excluded (0) from the subset.

beginner

How many subsets can be generated from a set of size n using bitmask?

There are 2^n subsets because each element can either be included or excluded, giving 2 choices per element.

beginner

In bitmask subset generation, what does the bit at position i represent?

The bit at position i indicates whether the element at index i in the original set is included (1) or excluded (0) in the current subset.

intermediate

Why is bitmask subset generation efficient for small sets?

Because it uses simple binary operations to generate all subsets without recursion, making it fast and easy to implement for sets with small size.

beginner

What is the output of subsets generated by bitmask for the set [a, b]?

The subsets are: [], [a], [b], [a, b]. These correspond to bitmasks 00, 01, 10, and 11 respectively.

How many subsets does a set with 3 elements have?

A 8

B 6

C 3

D 9

A set with 3 elements has 2^3 = 8 subsets.

In bitmask subset generation, what does a bit value of 0 mean for an element?

A Element is excluded from the subset

B Element is included in the subset

C Element is duplicated

D Element is the first in the subset

A bit value of 0 means the element is excluded from the subset.

Which operation is commonly used to check if the i-th bit is set in a bitmask?

A OR (|)

B XOR (^)

C AND (&)

D NOT (~)

The AND operation with (1 << i) checks if the i-th bit is set.

What is the bitmask for the subset including only the first and third elements of a 3-element set?

A 111 (binary)

B 110 (binary)

C 011 (binary)

D 101 (binary)

Bits 0 and 2 set means 101 in binary.

Why is bitmask generation not suitable for very large sets?

A Because it uses recursion

B Because the number of subsets grows exponentially

C Because it requires sorting

D Because it only works for sets of size 2

The number of subsets is 2^n, which grows very fast and becomes impractical for large n.

Explain how to generate all subsets of a set using bitmasking.

Think of binary numbers representing subsets.

Describe why the total number of subsets is 2^n for a set of size n.

Consider each element as a yes/no decision.

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