Count Number of Max Bitwise-OR Subsets

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Steps

setup

Initialize variables

Initialize n as the length of nums, max_or as 0, and count as 0 before starting subset enumeration.

💡 Setting up these variables is essential to track the maximum OR value and how many subsets achieve it.

Line:

n = len(nums)
max_or = 0
count = 0

💡 We start with no subsets considered, so max_or and count are zero.

traverse

Start subset enumeration with mask=1

Begin iterating subsets by setting mask to 1, representing the subset including only the first element.

💡 Each mask represents a subset; mask=1 means only the first element is included.

Line:

for mask in range(1, 1 << n):
    current_or = 0

💡 We will compute the OR for this subset next.

fill_cells

Compute OR for mask=1, include nums[0]

Check bit 0 of mask=1; since it is set, include nums[0]=3 in current_or by OR operation.

💡 Checking each bit of mask determines which elements are included in the subset.

Line:

for i in range(n):
    if mask & (1 << i):
        current_or |= nums[i]

💡 Bitwise AND with mask identifies included elements; OR accumulates their values.

fill_cells

Skip nums[1] for mask=1

Check bit 1 of mask=1; it is not set, so nums[1] is excluded from current_or.

💡 Bits not set in mask mean the corresponding elements are excluded from the subset.

Line:

for i in range(n):
    if mask & (1 << i):
        current_or |= nums[i]

💡 Only elements with bits set in mask contribute to the OR.

compare

Update max_or and count after mask=1

current_or is 3 for mask=1; since 3 > max_or(0), update max_or to 3 and reset count to 1.

💡 We track the highest OR value and how many subsets achieve it.

Line:

if current_or > max_or:
    max_or = current_or
    count = 1

💡 A new maximum OR subset is found, so count resets to 1.

traverse

Start subset enumeration with mask=2

Set mask to 2, representing the subset including only the second element nums[1]=1.

💡 Next subset to evaluate includes only the second element.

Line:

for mask in range(1, 1 << n):
    current_or = 0

💡 We will compute the OR for this subset next.

fill_cells

Skip nums[0] for mask=2

Check bit 0 of mask=2; it is not set, so nums[0] is excluded from current_or.

💡 Bit 0 not set means first element is excluded.

Line:

for i in range(n):
    if mask & (1 << i):
        current_or |= nums[i]

💡 Only elements with bits set in mask contribute to the OR.

fill_cells

Include nums[1] for mask=2

Check bit 1 of mask=2; it is set, so include nums[1]=1 in current_or.

💡 Bit 1 set means second element is included.

Line:

for i in range(n):
    if mask & (1 << i):
        current_or |= nums[i]

💡 OR accumulates included elements' values.

compare

Compare current_or with max_or after mask=2

current_or is 1 for mask=2; since 1 < max_or(3), max_or and count remain unchanged.

💡 Only update max_or and count if current_or is greater or equal to max_or.

Line:

if current_or > max_or:
    max_or = current_or
    count = 1
elif current_or == max_or:
    count += 1

💡 Subsets with OR less than max_or do not affect count.

traverse

Start subset enumeration with mask=3

Set mask to 3, representing the subset including both nums[0]=3 and nums[1]=1.

💡 This subset includes all elements.

Line:

for mask in range(1, 1 << n):
    current_or = 0

💡 We will compute the OR for this full subset next.

fill_cells

Include nums[0] for mask=3

Bit 0 of mask=3 is set, so include nums[0]=3 in current_or.

💡 First element is included in this subset.

Line:

for i in range(n):
    if mask & (1 << i):
        current_or |= nums[i]

💡 OR accumulates included elements.

fill_cells

Include nums[1] for mask=3

Bit 1 of mask=3 is set, so include nums[1]=1 in current_or.

💡 Second element is included in this subset.

Line:

for i in range(n):
    if mask & (1 << i):
        current_or |= nums[i]

💡 OR accumulates all included elements.

compare

Compare current_or with max_or after mask=3

current_or is 3 for mask=3; since 3 == max_or(3), increment count to 2.

💡 Subsets with OR equal to max_or increase the count.

Line:

if current_or > max_or:
    max_or = current_or
    count = 1
elif current_or == max_or:
    count += 1

💡 Multiple subsets can share the maximum OR value.

reconstruct

Return final count

All subsets processed; return count which is 2, the number of subsets with max OR.

💡 The final count is the answer to the problem.

Line:

return count

💡 The algorithm correctly counted subsets with maximum bitwise OR.

from typing import List

class Solution:
    def countMaxOrSubsets(self, nums: List[int]) -> int:
        n = len(nums)  # STEP 1
        max_or = 0    # STEP 1
        count = 0     # STEP 1
        for mask in range(1, 1 << n):  # STEP 2,6,10
            current_or = 0             # STEP 2,6,10
            for i in range(n):         # STEP 3,4,7,8,11,12
                if mask & (1 << i):   # STEP 3,4,7,8,11,12
                    current_or |= nums[i]  # STEP 3,8,11,12
            if current_or > max_or:    # STEP 5,13
                max_or = current_or    # STEP 5
                count = 1             # STEP 5
            elif current_or == max_or: # STEP 13
                count += 1            # STEP 13
        return count                 # STEP 14

if __name__ == '__main__':
    sol = Solution()
    print(sol.countMaxOrSubsets([3,1]))  # Expected output: 2
    print(sol.countMaxOrSubsets([2,2,2]))  # Expected output: 7

📊

Count Number of Max Bitwise-OR Subsets - Watch the Algorithm Execute, Step by Step

Watching each subset being generated and evaluated helps you understand how bitmask enumeration works and why the algorithm correctly counts subsets with the maximum bitwise OR.

enteringnums=[3,1]

Call Path (current → root)

countMaxOrSubsets([3,1])

Remaining

mask=1 to 3

choosingnums=[3,1]

Call Path (current → root)

countMaxOrSubsets([3,1])

Tried

mask=1

Remaining

mask=2mask=3

Partial: [mask=1]

choosingnums=[3,1]

Call Path (current → root)

countMaxOrSubsets([3,1])

Tried

mask=1, i=0 included

Remaining

mask=1, i=1 excluded

Partial: [mask=1, included nums[0]=3]

choosingnums=[3,1]

Call Path (current → root)

countMaxOrSubsets([3,1])

Tried

mask=1, i=0 includedmask=1, i=1 excluded

Partial: [mask=1, included nums[0]=3]

choosingnums=[3,1]

Call Path (current → root)

countMaxOrSubsets([3,1])

Tried

mask=1 processed

Remaining

mask=2mask=3

choosingnums=[3,1]

Call Path (current → root)

countMaxOrSubsets([3,1])

Tried

mask=1mask=2

Remaining

mask=3

Partial: [mask=2]

choosingnums=[3,1]

Call Path (current → root)

countMaxOrSubsets([3,1])

Tried

mask=2, i=0 excluded

Remaining

mask=2, i=1 included

Partial: [mask=2]

choosingnums=[3,1]

Call Path (current → root)

countMaxOrSubsets([3,1])

Tried

mask=2, i=0 excludedmask=2, i=1 included

Partial: [mask=2, included nums[1]=1]

choosingnums=[3,1]

Call Path (current → root)

countMaxOrSubsets([3,1])

Tried

mask=1mask=2

Remaining

mask=3

choosingnums=[3,1]

Call Path (current → root)

countMaxOrSubsets([3,1])

Tried

mask=1mask=2mask=3

Partial: [mask=3]

choosingnums=[3,1]

Call Path (current → root)

countMaxOrSubsets([3,1])

Tried

mask=3, i=0 included

Remaining

mask=3, i=1 included

Partial: [mask=3, included nums[0]=3]

choosingnums=[3,1]

Call Path (current → root)

countMaxOrSubsets([3,1])

Tried

mask=3, i=0 includedmask=3, i=1 included

Partial: [mask=3, included nums[0]=3, included nums[1]=1]

choosingnums=[3,1]

Call Path (current → root)

countMaxOrSubsets([3,1])

Tried

mask=1mask=2mask=3

backtrackingnums=[3,1]

Call Path (current → root)

countMaxOrSubsets([3,1])

Tried

mask=1mask=2mask=3

Found 1: [2]

Key Takeaways

✓ Enumerating subsets with bitmasks allows systematic and complete exploration of all subsets.

This is hard to see from code alone because the bitmask iteration looks like a loop, but visualizing each mask as a subset clarifies the process.

✓ The bitwise OR operation accumulates values from included elements, revealing how subsets combine bits.

Seeing the OR build step-by-step helps understand why the OR of a subset is computed by OR-ing included elements.

✓ Tracking max_or and count separately shows how the algorithm updates the best OR and counts subsets achieving it.

The decision to reset count or increment it is clearer when watching each subset's OR compared to max_or.

Practice

1. You are given a set of distinct positive integers and a target sum. You need to find all unique combinations where the chosen numbers sum to the target. Each number can be chosen multiple times. Which algorithmic approach best guarantees finding all valid combinations efficiently without missing any or producing duplicates?

easy

A. Greedy algorithm that picks the largest candidate repeatedly until the sum is reached or exceeded.

B. Backtracking with sorting and frequency counting to avoid duplicates and prune early when sum exceeds target.

C. Dynamic programming using a 2D table to count the number of combinations without generating them explicitly.

D. Pure brute force recursion that tries all subsets without pruning or sorting.

Solution

1. Step 1: Understand problem constraints

   The problem requires all unique combinations summing to target, allowing reuse of candidates.

2. Step 2: Evaluate algorithm suitability

   Greedy fails because it can miss combinations; pure brute force is correct but inefficient; DP counts combinations but doesn't generate them; backtracking with sorting and frequency counting avoids duplicates and prunes search space efficiently.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Backtracking with pruning and frequency counting is the standard approach [OK]

Hint: Backtracking with pruning avoids duplicates and excessive search [OK]

Common Mistakes:

• Assuming greedy always works for sum problems
• Confusing counting with generating combinations

2. Consider the following Python function that counts subsets with sum K using a space-optimized DP approach. What is the output when called with arr = [1, 2, 3, 3] and K = 6?

def count_subsets_space_optimized(arr, K):
    dp = [0] * (K + 1)
    dp[0] = 1
    for num in arr:
        for j in range(K, num - 1, -1):
            dp[j] += dp[j - num]
    return dp[K]

print(count_subsets_space_optimized(arr, K))

easy

A. 5

B. 4

C. 3

D. 6

Solution

1. Step 1: Initialize dp array

   dp = [1,0,0,0,0,0,0] since dp[0]=1 and K=6.

2. Step 2: Process each number updating dp

   After processing 1: dp = [1,1,0,0,0,0,0] After 2: dp = [1,1,1,1,0,0,0] After first 3: dp = [1,1,1,2,1,1,1] After second 3: dp = [1,1,1,3,2,2,3]

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   dp[6] = 5 subsets sum to 6 [OK]

Hint: Trace dp updates backward to avoid double counting [OK]

Common Mistakes:

• Forgetting to iterate backward in dp array
• Off-by-one errors in dp indices

3. Consider the following Python code implementing the optimal backtracking with bitmask and memoization approach for the Matchsticks to Square problem. Given the input [1,1,2,2,2], what is the final return value of makesquare([1,1,2,2,2])?

easy

A. True

B. Depends on the order of matchsticks

C. Raises an exception due to index error

D. False

Solution

1. Step 1: Calculate total and side length

   Total = 1+1+2+2+2 = 8, side = 8/4 = 2.

2. Step 2: Trace backtracking calls

   Sorted sticks: [2,2,2,1,1]. The algorithm tries to form sides of length 2. It can form three sides with three 2's, and the last side with two 1's. Hence, returns true.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Sum divisible by 4 and sticks fit sides exactly [OK]

Hint: Sum divisible by 4 and sticks fit sides exactly [OK]

Common Mistakes:

• Forgetting to sort and prune
• Assuming order affects result
• Miscounting sides formed

4. The following code attempts to solve Combination Sum (Reuse Allowed). Identify the line containing the subtle bug that causes incorrect results or inefficiency:

def combinationSum(candidates, target):
    candidates.sort()
    result = []
    def backtrack(index, path, target):
        if target == 0:
            result.append(path[:])
            return
        if index == len(candidates) or target < 0:
            return
        max_use = target // candidates[index]
        for count in range(max_use + 1):
            path += [candidates[index]] * count
            backtrack(index + 1, path, target - candidates[index] * count)
            # Missing path restoration here
    backtrack(0, [], target)
    return result

medium

A. Line with 'path += [candidates[index]] * count' because path is modified without backtracking (pop).

B. Line with 'candidates.sort()' because sorting is unnecessary.

C. Line with 'if target == 0:' because it should check for target <= 0.

D. Line with 'max_use = target // candidates[index]' because integer division may cause errors.

Solution

1. Step 1: Analyze path modification in loop

   Path is extended by count copies of candidates[index] but never restored after recursive call, causing path to grow incorrectly across iterations.

2. Step 2: Identify missing backtracking step

   Proper backtracking requires removing added elements after recursion to restore path state for next iteration.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Modifying path without restoration leads to incorrect combinations [OK]

Hint: Always restore path after recursion in backtracking [OK]

Common Mistakes:

• Forgetting to pop elements after recursion
• Assuming path copy avoids mutation issues

5. What is the time complexity of generating all subsets of an array of size n using bitmask enumeration, and why is the common misconception that it is O(2^n) incorrect?

medium

A. O(n^2) because we iterate over n elements twice

B. O(2^n) because there are 2^n subsets

C. O(n * 2^n) because each subset can have up to n elements and we build each subset by checking n bits

D. O(n) because we process each element once

Solution

1. Step 1: Count subsets and subset size

   There are 2^n subsets, and each subset can have up to n elements.

2. Step 2: Analyze bitmask enumeration cost

   For each of the 2^n masks, we check n bits to build the subset, resulting in O(n * 2^n) time.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Time depends on both number of subsets and subset size [OK]

Hint: Each subset requires O(n) to build, total O(n*2^n) [OK]

Common Mistakes:

• Ignoring subset size in complexity
• Assuming O(2^n) only
• Confusing with DP complexities