Largest Combination With Bitwise AND Greater Than Zero

Problem Statement

In the realm of binary arithmetic, the bitwise AND operation is a fundamental operation where two bits are compared, and the result is 1 only if both bits are 1, otherwise it is 0. When applied to a sequence of numbers, bitwise AND evaluates to 1 at each bit position where all the numbers in sequence also have 1 at the same bit position.

Given the context, you are tasked with an interesting challenge involving an array of positive integers named candidates. Your role is to compute the bitwise AND for all possible combinations of elements within the candidates array. Specifically, you need to determine the size of the largest combination where the resulting bitwise AND is greater than zero.

Illustrated with an example:

• For a set of numbers like candidates = [1, 5, 3], calculating the bitwise AND for all possible combinations, we would identify which combination yields the highest count of numbers while maintaining a non-zero result of the bitwise AND operation.
• The operation essentially filters down combinations based on their collective binary characteristics, focusing on finding the maximum grouping where all combined numbers share at least one common binary 1.

This leads us to return the count of elements in the largest valid grouping, where "valid" means their bitwise AND does result in a value greater than zero.

Examples

Example 1

Input:

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

Output:

4

Explanation:

The combination [16,17,62,24] has a bitwise AND of 16 & 17 & 62 & 24 = 16 > 0.
The size of the combination is 4.
It can be shown that no combination with a size greater than 4 has a bitwise AND greater than 0.
Note that more than one combination may have the largest size.
For example, the combination [62,12,24,14] has a bitwise AND of 62 & 12 & 24 & 14 = 8 > 0.

Example 2

Input:

candidates = [8,8]

Output:

2

Explanation:

The largest combination [8,8] has a bitwise AND of 8 & 8 = 8 > 0.
The size of the combination is 2, so we return 2.

Constraints

• 1 <= candidates.length <= 105
• 1 <= candidates[i] <= 107

Approach and Intuition

Understanding how the bitwise AND operation impacts the combination plays a crucial role in solving this problem. The challenge boils down to identifying the largest subset of the candidates array wherein the bitwise AND of its elements results in a value greater than zero. Here's a step-by-step insight into approaching the task:

1. Categorize bits positions:

   • Track which bit positions (0 through up to 30 for integers within the provided constraints) are commonly set to 1 across different elements.
   • This is done by looking at each bit position individually and counting its appearance across all elements in terms of it being set (1).

2. Optimize by most common bits:

   • Determine combinations starting from the most common bits, as these bits that are more frequently set across the array elements will be more likely to achieve a bitwise AND greater than zero in larger groups.

3. Calculate the size of largest combination for each significant bit:

   • For each possible bit (starting from the most significant to the least), group all numbers having that bit set.

4. Determine the maximum size:

   • The size of the largest group formed around a significant 1 bit gives the answer. This is because members of such a group, when ANDed together, will naturally yield a result with at least that one 1 bit set, thus greater than zero.

The result derived from the above steps offers not just the solution but insight into how binary operations can determine and influence number groupings based on shared bit characteristics. This process taps deeply into binary arithmetic principles, specifically looking at communalities and variations across the digital spectrum of given numbers.

Solutions

class Solution {
public:
    int maxSetBits(vector<int>& nums) {
        int highestBitCount = 0;
        for (int i = 0; i < 24; ++i) {
            int bitCount = 0;
            for (int number : nums) {
                if (number & (1 << i)) {
                    bitCount++;
                }
            }
            highestBitCount = max(highestBitCount, bitCount);
        }
        return highestBitCount;
    }
};