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2275. Largest Combination With Bitwise AND Greater Than Zero

Description

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

Solutions

Solution 1: Bit Manipulation

The problem requires finding the maximum length of a combination of numbers where the bitwise AND result is greater than \(0\). This implies that there must be a certain binary bit where all numbers have a \(1\) at that position. Therefore, we can enumerate each binary bit and count the number of \(1\)s at that bit position for all numbers. Finally, we take the maximum count.

The time complexity is \(O(n \times \log M)\), where \(n\) and \(M\) are the length of the array \(\textit{candidates}\) and the maximum value in the array, respectively. The space complexity is \(O(1)\).

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class Solution:
    def largestCombination(self, candidates: List[int]) -> int:
        ans = 0
        for i in range(max(candidates).bit_length()):
            ans = max(ans, sum(x >> i & 1 for x in candidates))
        return ans

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class Solution {
    public int largestCombination(int[] candidates) {
        int mx = Arrays.stream(candidates).max().getAsInt();
        int m = Integer.SIZE - Integer.numberOfLeadingZeros(mx);
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            int cnt = 0;
            for (int x : candidates) {
                cnt += x >> i & 1;
            }
            ans = Math.max(ans, cnt);
        }
        return ans;
    }
}

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class Solution {
public:
    int largestCombination(vector<int>& candidates) {
        int mx = *max_element(candidates.begin(), candidates.end());
        int m = 32 - __builtin_clz(mx);
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            int cnt = 0;
            for (int x : candidates) {
                cnt += x >> i & 1;
            }
            ans = max(ans, cnt);
        }
        return ans;
    }
};

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func largestCombination(candidates []int) (ans int) {
    mx := slices.Max(candidates)
    m := bits.Len(uint(mx))
    for i := 0; i < m; i++ {
        cnt := 0
        for _, x := range candidates {
            cnt += (x >> i) & 1
        }
        ans = max(ans, cnt)
    }
    return
}

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function largestCombination(candidates: number[]): number {
    const mx = Math.max(...candidates);
    const m = mx.toString(2).length;
    let ans = 0;
    for (let i = 0; i < m; ++i) {
        let cnt = 0;
        for (const x of candidates) {
            cnt += (x >> i) & 1;
        }
        ans = Math.max(ans, cnt);
    }
    return ans;
}

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