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Problem 2275 Largest Combination With Bitwise AND Greater Than Zero

Problem Statement

Link - Problem 2275

Question

The bitwise AND of an array nums is the bitwise AND of all integers in nums.

• For example, for nums = [1, 5, 3], the bitwise AND is equal to 1 & 5 & 3 = 1.
• Also, for nums = [7], the bitwise AND is 7.

You are given an array of positive integers candidates. Compute the bitwise AND for all possible combinations of elements in the candidates array.

Return the size of the largest combination of candidates with a bitwise AND greater than 0.

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

Solution

class Solution {
public:
    int largestCombination(vector<int>& candidates) {
        vector<int>bit_count(24,0);
        for(int i=0;i<candidates.size();i++){
            int temp = candidates[i];
            int j=0;

            while(temp>0&&j<24){
                if((temp&1)==1){
                    bit_count[j]++;
                }
                j++;
                temp= temp>>1;
            }
        }

        int max_count = INT_MIN;
        for(auto it:bit_count)
            max_count = max(max_count,it);

        return max_count;
    }
};

Complexity Analysis

| Algorithm         | Time Complexity | Space Complexity |
| ----------------- | --------------- | ---------------- |
| Bitwise Iteration | O(n log n)      | O(1)             |

Explanation

Intial Thoughts

To start solving this problem, consider the properties of bitwise AND operations, such as the fact that it returns 1 only if both bits are 1. Next, think about how to represent the given numbers in binary form and how their bitwise AND would look. It’s also essential to consider how to generate all possible combinations of elements in the candidates array. Think about the concept of a bit count and how it can help in solving this problem. Consider the time complexity and how to optimize it. Reflect on the fact that we are looking for the largest combination, so we need to find the maximum size of a combination with a bitwise AND greater than 0.

Intuitive Analysis

The solution to this problem involves counting the number of 1s in each bit position across all numbers in the array. To intuitively solve this problem, think of each bit position as a separate ‘slot’ where we count how many numbers have a 1 in that position. Then, look for the slot with the highest count because that’s the bit position where the most numbers have a 1, indicating the largest combination with a bitwise AND greater than 0. Analyze how the while loop and the bitwise operations work together to achieve this. Consider the idea that if a bit position has the maximum count, it means that all numbers in the largest combination have a 1 in that position, resulting in a bitwise AND greater than 0 for that combination.

1. Intuition

• The problem requires finding the largest combination of elements in the candidates array with a bitwise AND greater than 0.
• To solve this, we need to understand how bitwise AND operations work and how they can be used to find the largest combination.
• We can use a bit count array to keep track of the count of each bit in the candidates array.
• The maximum count in the bit count array will give us the size of the largest combination with a bitwise AND greater than 0.
• This approach works because the bitwise AND of two numbers will have a bit set to 1 only if the corresponding bits in both numbers are 1.
• By counting the number of bits set to 1 in each position, we can find the maximum number of elements that can be combined to get a bitwise AND greater than 0.
• This solution has a time complexity of O(n) and a space complexity of O(1), where n is the number of elements in the candidates array.

2. Implementation

• We initialize a bit_count array of size 24 to keep track of the count of each bit in the candidates array.
• We iterate over each element in the candidates array and use a while loop to count the number of bits set to 1 in each position using the temp & 1 and temp >> 1 operations.
• We update the bit_count array accordingly using the bit_count[j]++ operation.
• After counting the bits for all elements, we find the maximum count in the bit_count array using the max function.
• We return the maximum count as the size of the largest combination with a bitwise AND greater than 0.
• The INT_MIN variable is used to initialize the max_count variable to negative infinity, ensuring that the first count will be greater than it.
• The max function is used to update the max_count variable with the maximum count found so far.

Complexity Analysis

Time Complexity:

• The algorithm iterates over each candidate in the input vector, and for each candidate, it performs a bitwise iteration using temp = temp >> 1, resulting in a time complexity of O(n log m), where n is the number of candidates and m is the maximum value in the candidates vector. However, since m is constrained to be at most 2^24 - 1 (as indicated by the size of the bit_count vector), we can consider log m to be a constant, making the time complexity O(n).
• The dominant operation is the while loop with the condition temp > 0 && j < 24, which iterates over each bit in the binary representation of temp. The loop runs in O(log m) time, where m is the value of temp.
• Considering the outer loop iterates over each candidate in O(n) time and the inner loop runs in O(log m) time, the overall time complexity can be classified as O(n) due to the constraint on the size of m.

Space Complexity:

• The algorithm uses a vector bit_count of size 24 to store the count of each bit position across all candidates, resulting in a space complexity of O(1) because the size of the vector is constant.
• The space usage does not grow with the size of the input vector candidates, making the space complexity O(1).
• The use of a constant-size vector bit_count justifies the O(1) space complexity, as it does not depend on the input size.

Footnote

This question is rated as Medium difficulty.

Hints

For the bitwise AND to be greater than zero, at least one bit should be 1 for every number in the combination.

The candidates are 24 bits long, so for every bit position, we can calculate the size of the largest combination such that the bitwise AND will have a 1 at that bit position.

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