In this problem, we’re exploring bitwise operations, particularly the bitwise AND operation. Our goal is to find the largest subset of numbers from a list such that the bitwise AND of all numbers in the subset is greater than zero. 🔥

📜 Problem Statement

We are given an array of positive integers, candidates. Our task is to evaluate every possible combination of these integers and determine the largest subset for which the bitwise AND is greater than 0. Each number in candidates can only be used once per combination.

Let’s break down the requirements and the bitwise operations involved! 🚀

🧩 Example Walkthrough

Example 1

Input: candidates = [16, 17, 71, 62, 12, 24, 14]
Output: 4
Explanation: The largest subset with a bitwise AND > 0 could be [16, 17, 62, 24] where the bitwise AND calculation yields a positive result. No larger combination exists that maintains this property.

Example 2

Input: candidates = [8, 8]
Output: 2
Explanation: Here, both values are the same, and combining them results in a bitwise AND of 8, which is positive. Therefore, the largest subset with a bitwise AND > 0 has size 2.

🧠 Understanding the Approach and Edge Cases

Edge Cases to Consider 🛠️

1. Single Element: When candidates has only one element (e.g., [5]). Here, the answer is 1 because the element itself is the only valid subset.
2. All Elements Are Power of Two: Special case where numbers like [1, 2, 4, 8] are considered. Each only has one bit set to 1, so the bitwise AND of any combination of them would be 0 unless repeated.
3. High Number of Duplicates: If all numbers are the same, such as [8, 8, 8, 8], the answer is simply the length of the array because the bitwise AND will always yield a positive result.
4. Very Large Array with Small Values: If there’s a large number of small values, such as [1, 1, 1, ...] (up to max length), we should ensure our solution efficiently handles this without redundant calculations.

🚀 Solution Overview

The bitwise AND operation only results in a non-zero value if all numbers in the subset have at least one common bit set to 1. Thus, for every bit position (up to 24 bits since 1 <= candidates[i] <= 10^7), we can:

1. Count how many numbers in candidates have a 1 at each bit position.
2. For each bit position, the count of numbers with 1 at that position represents the size of the largest subset with a positive bitwise AND result when restricted to that bit.

Why This Works:

If a bit position has many numbers with 1, these numbers can be combined, ensuring the bitwise AND result at that position remains greater than zero.

Steps:

1. Iterate through each bit position (up to 24 bits).
2. Count how many numbers have 1 in each position.
3. Track the maximum count across all bit positions, as this count will represent the size of the largest subset with a positive AND result.

💡 Optimized Solution

This optimized approach leverages bitwise operations efficiently by focusing only on the bits that matter rather than generating all combinations.

Python Code Implementation

Here’s how this solution would look in Python:

def largestCombination(candidates):
    max_size = 0
    
    # Check each bit position up to 24 bits (sufficient for numbers up to 10^7)
    for bit in range(24):
        count = 0
        # Count how many numbers have '1' at the current bit position
        for num in candidates:
            if num & (1 << bit):
                count += 1
        # Update max_size if the current count is greater
        max_size = max(max_size, count)
    
    return max_size

🕹️ Explanation of the Code

1. Bit Iteration: We loop through each of the 24 possible bit positions, as a 24-bit integer can represent any number up to the problem’s limit.
2. Counting: For each position, we count how many numbers in candidates have 1 at the current position using num & (1 << bit).
3. Updating Maximum: The variable max_size keeps track of the largest subset size with a bitwise AND > 0.

⏱️ Time Complexity Analysis

• Time Complexity: O(24 * n), where n is the length of candidates. This is efficient, given that 24 is a constant and n can be very large.
• Space Complexity: O(1), as we use only a fixed number of additional variables.

🖼️ Example Walkthrough: Detailed Calculation with Larger Numbers

Let’s go through Example 1 with a detailed bit-level walkthrough to reinforce understanding.

Example: candidates = [16, 17, 71, 62, 12, 24, 14]

1. Bit Representation:
   • 16 -> 00010000
   • 17 -> 00010001
   • 71 -> 01000111
   • 62 -> 00111110
   • 12 -> 00001100
   • 24 -> 00011000
   • 14 -> 00001110
2. Counting 1s at Each Bit Position:
   • Bit position 0: 17, 71, 62, 14 → 4 numbers have 1 here.
   • Bit position 4: 16, 17, 62, 24 → 4 numbers have 1 here.
   • …
3. Result: The largest count across all bit positions is 4, so the output is 4.

🌟 Edge Case Walkthroughs

1. Single Element:
   • Input: [1]
   • Output: 1
   • Explanation: Since there’s only one element, it forms the only valid subset.
2. Duplicate Values:
   • Input: [4, 4, 4, 4]
   • Output: 4
   • Explanation: Here, all values are identical, so we can combine all four to get a subset size of 4.
3. Mixed Large and Small Values:
   • Input: [1, 2, 4, 8, 16]
   • Output: 1
   • Explanation: Each value has only one unique bit set to 1, so no combination will have a positive bitwise AND result.

📊 Conclusion

In this problem, we explored how to use bitwise operations to solve a combinatorial problem efficiently! Instead of generating combinations, we focused on bitwise positions and identified the largest subset for each position that ensures a positive bitwise AND result. 🥳

This approach highlights how efficient thinking can transform complex problems into manageable solutions by focusing on what actually matters for a successful result—in this case, individual bit positions. 💪