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2275 - Largest Combination With Bitwise AND Greater Than Zero (Medium)

https://leetcode.com/problems/largest-combination-with-bitwise-and-greater-than-zero

Problem Statement​

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. Evaluate the bitwise AND of every combination of numbers of candidates. Each number in candidates may only be used once in each combination.

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 <= 10^5
  • 1 <= candidates[i] <= 10^7

Approach 1: Counting​

If the ii-th bit is set after performing AND operation on kk candidates, it means the ii-th of all kk candidates must be 1. Therefore, we check each bit to see which bit has the most candidates with the bit set.

Written by @wingkwong
class Solution {
public:
    int largestCombination(vector<int>& candidates) {
        int n = candidates.size(), ans = 0;
        for (int i = 0; i < 32; i++) {
            // init bits to  0 for each index
            int bits = 0;
            // iterate each candidate
            for (int j = 0; j < n; j++) {
                // i-th bit is set for candidates[j]
                if ((1 << i) & candidates[j]) {
                    // count the bit
                    bits += 1;
                }
            }
            // update the ans 
            ans = max(ans, bits);
        }
        return ans;
    }
};