Subsets Generation Using Bitmask in DSA Go - 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.

intermediate

In Go, how do you check if the j-th bit in an integer i is set?

Use the expression (i & (1 << uint(j))) != 0 to check if the j-th bit is 1.

intermediate

Why is bitmask method efficient for generating subsets compared to recursion?

Bitmask uses simple bit operations and loops, avoiding function call overhead and making it easy to implement iteratively.

beginner

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

The subsets are: [], [a], [b], [a, b].

How many subsets does a set with 3 elements have?

A 8

B 3

C 9

D 6

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

Which bitwise operation checks if a bit is set in bitmask?

A XOR (^)

B AND (&)

C OR (|)

D NOT (~)

AND (&) operation with a mask checks if a specific bit is set.

What does the bitmask 0101 represent for a set [a, b, c, d]?

A Subset with elements a and d

B Subset with elements b and d

C Subset with elements a and c

D Subset with elements b and c

Bits 0 and 2 are set (counting from 0), so elements a and c are included.

What is the time complexity of generating all subsets using bitmask for n elements?

A O(2^n)

B O(n)

C O(n^2)

D O(log n)

There are 2^n subsets, so the time complexity is O(2^n).

In Go, what does (1 << j) do?

A Performs bitwise OR with j

B Shifts 1 to the right by j bits

C Performs bitwise AND with j

D Shifts 1 to the left by j bits

(1 << j) shifts the number 1 to the left by j bits, creating a mask with only the j-th bit set.

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

Think of each subset as a number where bits tell which elements to pick.

Describe how to check if an element should be included in a subset using bitmask in Go.

Use (i & (1 << uint(j))) != 0 to check the j-th bit.

Practice

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

easy

A. Only subsets with odd number of elements

B. Only the largest subset

C. All possible subsets of the set

D. Only subsets with even number of elements

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 all bitmasks

   By iterating from 0 to 2^n - 1, all combinations of included/excluded elements are covered, producing all subsets.

3. Final Answer:

   All possible subsets of the set -> Option C

4. Quick Check:

   Bitmask = subsets generator [OK]

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

Common Mistakes:

• Thinking bitmask generates only specific subsets
• Confusing bitmask with sorting or filtering
• Ignoring subsets with zero elements

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

easy

A. for mask := 0; mask < n; mask++ { }"

B. for mask := 1; mask <= n; mask++ { }"

C. for mask := 0; mask <= n; mask++ { }"

D. for mask := 0; mask < (1 << n); mask++ { }"

Solution

1. Step 1: Understand bitmask range

   For n elements, bitmasks range from 0 to 2^n - 1, which is (1 << n) - 1.

2. Step 2: Correct loop condition

   The loop must run from 0 up to but not including (1 << n), so mask < (1 << n) is correct.

3. Final Answer:

   for mask := 0; mask < (1 << n); mask++ { } -> Option D

4. Quick Check:

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

Hint: Use mask < (1 << n) for full bitmask range [OK]

Common Mistakes:

• Using mask <= n instead of mask < (1 << n)
• Starting mask from 1 misses empty subset
• Using mask < n only iterates n times, not 2^n

3. What is the output subsets generated by the following Go code snippet?

arr := []int{1, 2}
for mask := 0; mask < (1 << 2); mask++ {
  subset := []int{}
  for i := 0; i < 2; i++ {
    if mask&(1<

medium

A. [ ]
[1]
[2]
[1 2]

B. [ ]
[2]
[1]
[1 2]

C. [1]
[2]
[1 2]
[ ]

D. [1 2]
[1]
[2]
[ ]

Solution

1. Step 1: Understand bitmask values

   For n=2, masks go 0 to 3: 0 (00), 1 (01), 2 (10), 3 (11).

2. Step 2: Map bits to subsets

   Mask 0: no bits set -> [] Mask 1: bit 0 set -> [1] Mask 2: bit 1 set -> [2] Mask 3: bits 0 and 1 set -> [1, 2]

3. Final Answer:

   [ ] [1] [2] [1 2] -> Option A

4. Quick Check:

   Bitmask order matches subsets [OK]

Hint: Map bits to elements in order from least significant bit [OK]

Common Mistakes:

• Mixing order of subsets printed
• Confusing which bit corresponds to which element
• Missing empty subset when mask=0

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

arr := []int{1, 2, 3}
for mask := 0; mask <= (1 << len(arr)); mask++ {
  subset := []int{}
  for i := 0; i < len(arr); i++ {
    if mask&(1<

medium

A. Loop condition should be mask < (1 << len(arr))

B. Bitwise AND operator should be | instead of &

C. Inner loop should run from i = 1 to len(arr)

D. Array indexing should start from 1 instead of 0

Solution

1. Step 1: Analyze loop condition

   The outer loop runs mask from 0 to (1 << n) inclusive, which is one too many iterations (2^n + 1).

2. Step 2: Correct loop range

   It should run from 0 up to but not including (1 << n), so mask < (1 << n) is correct to avoid out-of-range subsets.

3. Final Answer:

   Loop condition should be mask < (1 << len(arr)) -> Option A

4. Quick Check:

   Loop must run 2^n times exactly [OK]

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

Common Mistakes:

• Using <= causes one extra invalid iteration
• Confusing & with | for bit checking
• Changing loop indices incorrectly

5. Given a set arr := []int{1, 2, 3, 4}, how many subsets generated by bitmask will have exactly 2 elements?

hard

A. 4

B. 6

C. 8

D. 12

Solution

1. Step 1: Understand subsets count by size

   Number of subsets of size k from n elements is combination n choose k = C(n, k).

2. Step 2: Calculate C(4, 2)

   C(4, 2) = 4! / (2! * (4-2)!) = (4*3)/(2*1) = 6.

3. Final Answer:

   6 -> Option B

4. Quick Check:

   Number of 2-element subsets = 6 [OK]

Hint: Use combinations formula C(n, k) for subset sizes [OK]

Common Mistakes:

• Counting all subsets instead of only size 2
• Using permutations instead of combinations
• Misapplying formula for combinations