Subsets Generation Using Bitmask in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to calculate the total number of subsets for a set of size n.

DSA C++

int total_subsets = 1 << [1];

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

Asize

Bn

Ccount

Dlength

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong variable name instead of n.

Using addition instead of bit shift.

Using bit shifting 1 << n calculates 2 to the power n, which is the total number of subsets.

2fill in blank

medium

Complete the loop to iterate over all subsets represented by bitmasks.

DSA C++

for (int mask = 0; mask < (1 << [1]); mask++) {
    // process subset
}

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

Asize

Blength

Ccount

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong variable for the set size.

Using <= instead of < in the loop condition.

The loop runs from 0 to 2^n - 1, covering all subsets represented by bitmasks.

3fill in blank

hard

Fix the error in the condition to check if the j-th element is in the current subset mask.

DSA C++

if (mask & (1 << [1])) {
    subset.push_back(nums[j]);
}

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

Aj

Bi

Cmask

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using i instead of j for the bit position.

Using mask or n which are not bit positions.

We check the j-th bit in mask to see if the j-th element is included in the subset.

4fill in blank

hard

Fill both blanks to complete the inner loop that builds each subset from the bitmask.

DSA C++

for (int j = 0; j < [1]; j++) {
    if (mask & (1 << [2])) {
        subset.push_back(nums[j]);
    }
}

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

An

Bi

Cj

Dsize

Attempts:

3 left

💡 Hint

Common Mistakes

Using i instead of j in the bit check.

Using wrong loop limit.

The inner loop runs from 0 to n-1, and checks the j-th bit in mask to decide if nums[j] is included.

5fill in blank

hard

Fill all three blanks to complete the function that generates all subsets using bitmask.

DSA C++

vector<vector<int>> subsets(vector<int>& nums) {
    int [1] = nums.size();
    vector<vector<int>> result;
    for (int mask = 0; mask < (1 << [2]); mask++) {
        vector<int> subset;
        for (int j = 0; j < [3]; j++) {
            if (mask & (1 << j)) {
                subset.push_back(nums[j]);
            }
        }
        result.push_back(subset);
    }
    return result;
}

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

An

Bsize

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names inconsistently.

Using wrong loop limits.

We store the size in n, use n for the loop limit and bitmask range to generate all subsets.

Practice

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

easy

A. Only the largest subset of the set

B. All possible subsets of the set

C. The sum of all elements in the set

D. The sorted version 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 over all bitmasks from 0 to 2^n - 1, we get all possible subsets of the set.

3. Final Answer:

   All possible subsets of the set -> Option B

4. Quick Check:

   Bitmask = subsets generator [OK]

Hint: Bitmask bits map directly to subset elements included [OK]

Common Mistakes:

• Thinking bitmask generates only one subset
• Confusing subsets with permutations
• Assuming bitmask sorts elements

2. Which of the following is the correct C++ syntax to iterate over all bitmasks for a set of size n?

easy

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

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

C. for (int mask = 0; mask < (1 << n); mask++)

D. for (int mask = 0; mask < n; mask++)

Solution

1. Step 1: Understand bitmask range

   Bitmask ranges from 0 to 2^n - 1, which is (1 << n) - 1 in C++.

2. Step 2: Correct loop syntax

   The loop must run from 0 to less than (1 << n) to cover all subsets.

3. Final Answer:

   for (int mask = 0; mask < (1 << n); mask++) -> Option C

4. Quick Check:

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

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

Common Mistakes:

• Using mask <= n instead of mask < (1 << n)
• Starting loop from 1 instead of 0
• Using mask < n which is too small

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

std::vector<int> nums = {1, 2};
for (int mask = 0; mask < (1 << 2); mask++) {
    std::cout << "Subset: ";
    for (int i = 0; i < 2; i++) {
        if (mask & (1 << i)) std::cout << nums[i] << " ";
    }
    std::cout << std::endl;
}

medium

A. Subset:
Subset: 1
Subset: 2
Subset: 1 2

B. Subset: 1 2
Subset: 1
Subset: 2
Subset:

C. Subset: 1
Subset: 2
Subset: 1 2
Subset:

D. Subset:
Subset: 2
Subset: 1
Subset: 1 2

Solution

1. Step 1: Identify bitmask values and subsets

   For n=2, masks go 0 to 3:
   0 (00) no elements,
   1 (01) includes nums[0]=1,
   2 (10) includes nums[1]=2,
   3 (11) includes nums[0]=1 and nums[1]=2.

2. Step 2: Match output lines to subsets

   Output lines print subsets in order:
   "Subset: " (empty),
   "Subset: 1 ",
   "Subset: 2 ",
   "Subset: 1 2 "

3. Final Answer:

   Subset: Subset: 1 Subset: 2 Subset: 1 2 -> Option A

4. Quick Check:

   Bitmask 0 to 3 prints subsets in order [OK]

Hint: Map bits to elements; 0 means exclude, 1 means include [OK]

Common Mistakes:

• Mixing order of subsets
• Printing elements in wrong bit order
• Skipping empty subset output

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

std::vector<int> nums = {1, 2, 3};
for (int mask = 0; mask <= (1 << nums.size()); mask++) {
    for (int i = 0; i < nums.size(); i++) {
        if (mask & (1 << i)) std::cout << nums[i] << " ";
    }
    std::cout << std::endl;
}

medium

A. The vector nums should be declared as std::list instead.

B. The inner loop should run from i = 1 to nums.size().

C. The bitwise AND operator & should be replaced with &&.

D. The loop condition should be mask < (1 << nums.size()), not <=.

Solution

1. Step 1: Analyze loop range for bitmask

   The bitmask must go from 0 to 2^n - 1, so mask < (1 << n). Using <= causes one extra iteration beyond valid range.

2. Step 2: Check other parts

   Inner loop from 0 is correct, bitwise AND & is correct, vector type is fine.

3. Final Answer:

   The loop condition should be mask < (1 << nums.size()), not <=. -> Option D

4. Quick Check:

   Loop must run 0 to 2^n-1 [OK]

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

Common Mistakes:

• Using <= instead of < in loop condition
• Replacing bitwise & with logical &&
• Starting inner loop at 1 instead of 0

5. Given a set nums = {1, 2, 2}, which code modification correctly generates all unique subsets using bitmask technique in C++?

hard

A. Use a std::set<std::vector<int>> to store subsets and sort each subset before insertion.

B. Use bitmask from 1 to (1 << n) and print subsets directly without sorting.

C. Use a std::map<int, int> to count elements and generate subsets.

D. Skip duplicates by checking if current element equals previous without sorting.

Solution

1. Step 1: Understand duplicate issue in subsets

   With duplicates in input, bitmask generates duplicate subsets. Sorting subsets before storing helps identify duplicates.

2. Step 2: Use std::set to store unique subsets

   Storing sorted subsets in a set automatically removes duplicates, ensuring unique subsets only.

3. Final Answer:

   Use a std::set<std::vector<int>> to store subsets and sort each subset before insertion. -> Option A

4. Quick Check:

   Sort subsets + set = unique subsets [OK]

Hint: Sort subsets and use set to avoid duplicates [OK]

Common Mistakes:

• Not sorting subsets before inserting in set
• Using map instead of set for uniqueness
• Ignoring duplicates without sorting