Subsets Generation Using Bitmask in DSA Typescript - Build from Scratch

Subsets Generation Using Bitmask

📖 Scenario: You are helping a friend organize their collection of unique stickers. They want to see all possible groups of stickers they can create from their collection.

🎯 Goal: Build a program that generates all possible subsets (groups) of a given list of stickers using the bitmask technique.

📋 What You'll Learn

Create an array called stickers with the exact values ['star', 'heart', 'smile'].

Create a variable called n that stores the length of the stickers array.

Use a for loop with variable bitmask from 0 to (1 << n) - 1 to generate subsets.

Inside the loop, use another for loop with variable j from 0 to n - 1 to check bits.

Create an array called subset to hold the current subset stickers.

Print the array allSubsets containing all subsets after generation.

💡 Why This Matters

🌍 Real World

Generating all combinations of items is useful in planning, testing, and decision making.

💼 Career

Understanding subsets and bitmasking helps in coding interviews and solving combinatorial problems efficiently.

1

Create the stickers array

Create an array called stickers with these exact values: ['star', 'heart', 'smile'].

DSA Typescript

// Create the stickers array
// Your code here

Hint

Use const stickers = ['star', 'heart', 'smile']; to create the array.

2

Create the length variable

Create a variable called n that stores the length of the stickers array.

DSA Typescript

const stickers = ['star', 'heart', 'smile'];
// Create the variable n to store the length of stickers
// Your code here

Hint

Use const n = stickers.length; to get the number of stickers.

3

Generate all subsets using bitmask

Create an empty array called allSubsets. Use a for loop with variable bitmask from 0 to (1 << n) - 1. Inside it, create an empty array called subset. Use another for loop with variable j from 0 to n - 1. If the bit at position j in bitmask is set, add stickers[j] to subset. After the inner loop, add subset to allSubsets.

DSA Typescript

const stickers = ['star', 'heart', 'smile'];
const n = stickers.length;
// Generate all subsets using bitmask
// Your code here

Hint

Use two nested loops and bitwise AND to check bits. Add subsets to allSubsets.

4

Print all subsets

Print the array allSubsets to show all generated subsets.

DSA Typescript

const stickers = ['star', 'heart', 'smile'];
const n = stickers.length;
const allSubsets: string[][] = [];
for (let bitmask = 0; bitmask < (1 << n); bitmask++) {
    const subset: string[] = [];
    for (let j = 0; j < n; j++) {
        if ((bitmask & (1 << j)) !== 0) {
            subset.push(stickers[j]);
        }
    }
    allSubsets.push(subset);
}
// Print allSubsets
// Your code here

Hint

Use console.log(allSubsets); to print the result.

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