Subsets Generation Using Bitmask in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of subsets generated by bitmask

What is the output of the following C++ code that generates all subsets of {1, 2} using bitmask?

DSA C++

#include <iostream>
#include <vector>
using namespace std;

int main() {
    vector<int> nums = {1, 2};
    int n = nums.size();
    for (int mask = 0; mask < (1 << n); ++mask) {
        cout << "{";
        for (int i = 0; i < n; ++i) {
            if (mask & (1 << i)) {
                cout << nums[i] << " ";
            }
        }
        cout << "}\n";
    }
    return 0;
}

A

{}
{2 }
{1 }
{1 2 }

B

{}
{1 }
{2 }
{1 2 }

C

{1 }
{2 }
{1 2 }
{}

D

{1 2 }
{1 }
{2 }
{}

💡 Hint

Remember that bitmask from 0 to 2^n - 1 represents subsets, 0 means empty subset.

The bitmask loops from 0 to 3 (2^2 - 1). For each bit set, the corresponding element is included. mask=0 means empty set, mask=1 means {1}, mask=2 means {2}, mask=3 means {1,2}.

🧠 Conceptual

intermediate

Number of subsets generated by bitmask

If you have a set with 5 elements, how many subsets will the bitmask method generate?

A 10

B 25

C 32

D 64

💡 Hint

Number of subsets of a set with n elements is 2^n.

Each element can be either included or excluded, so total subsets = 2^5 = 32.

🔧 Debug

advanced

Identify the error in subset generation code

What error will the following code produce when generating subsets using bitmask for vector nums = {1, 2, 3}? int n = nums.size(); for (int mask = 0; mask <= (1 << n); ++mask) { for (int i = 0; i <= n; ++i) { if (mask & (1 << i)) { cout << nums[i] << " "; } } cout << "\n";

A Runtime error: out of range access

B Syntax error

C Infinite loop

D Runs fine and prints all subsets

💡 Hint

Check the loop condition for mask and how many subsets are generated.

The loop runs mask from 0 to (1 << n) inclusive, which is 2^n + 1 iterations. The last mask value equals 2^n, which has a bit set at position n, causing out of range access in nums[i].

❓ Predict Output

advanced

Output of subsets with sum calculation

What is the output of the following code that prints subsets of {1, 3} and their sums using bitmask?

DSA C++

#include <iostream>
#include <vector>
using namespace std;

int main() {
    vector<int> nums = {1, 3};
    int n = nums.size();
    for (int mask = 0; mask < (1 << n); ++mask) {
        int sum = 0;
        cout << "Subset: {";
        for (int i = 0; i < n; ++i) {
            if (mask & (1 << i)) {
                cout << nums[i] << " ";
                sum += nums[i];
            }
        }
        cout << "} Sum: " << sum << "\n";
    }
    return 0;

A

Subset: {} Sum: 0
Subset: {1 } Sum: 1
Subset: {3 } Sum: 3
Subset: {1 3 } Sum: 4

B

Subset: {} Sum: 1
Subset: {1 } Sum: 3
Subset: {3 } Sum: 4
Subset: {1 3 } Sum: 0

C

Subset: {1 } Sum: 1
Subset: {3 } Sum: 3
Subset: {1 3 } Sum: 4
Subset: {} Sum: 0

D

Subset: {3 } Sum: 3
Subset: {1 } Sum: 1
Subset: {1 3 } Sum: 4
Subset: {} Sum: 0

💡 Hint

Bitmask from 0 to 3 generates subsets in order: empty, {1}, {3}, {1,3}.

mask=0 no bits set sum=0, mask=1 includes 1 sum=1, mask=2 includes 3 sum=3, mask=3 includes 1 and 3 sum=4.

🧠 Conceptual

expert

Bitmask subset generation with duplicate elements

Given a vector nums = {1, 2, 2}, how many subsets will the bitmask method generate?

A 6

B 7

C 5

D 8

💡 Hint

Bitmask generates 2^3 = 8 subsets even with duplicate elements.

Bitmask generates 2^3=8 subsets including duplicates. Unique subsets are fewer but bitmask does not filter duplicates.

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