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3688. Bitwise OR of Even Numbers in an Array

Description

You are given an integer array nums.

Return the bitwise OR of all even numbers in the array.

If there are no even numbers in nums, return 0.

 

Example 1:

Input: nums = [1,2,3,4,5,6]

Output: 6

Explanation:

The even numbers are 2, 4, and 6. Their bitwise OR equals 6.

Example 2:

Input: nums = [7,9,11]

Output: 0

Explanation:

There are no even numbers, so the result is 0.

Example 3:

Input: nums = [1,8,16]

Output: 24

Explanation:

The even numbers are 8 and 16. Their bitwise OR equals 24.

 

Constraints:

  • 1 <= nums.length <= 100
  • 1 <= nums[i] <= 100

Solutions

Solution 1: Simulation

We define a variable \(\textit{ans}\) with an initial value of 0. Then, we iterate through each element \(x\) in the array \(\textit{nums}\); if \(x\) is even, we update \(\textit{ans}\) with the bitwise OR of \(\textit{ans}\) and \(x\).

Finally, we return \(\textit{ans}\).

The time complexity is \(O(n)\), where \(n\) is the length of the array \(\textit{nums}\). The space complexity is \(O(1)\).

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class Solution:
    def evenNumberBitwiseORs(self, nums: List[int]) -> int:
        return reduce(or_, (x for x in nums if x % 2 == 0), 0)

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class Solution {
    public int evenNumberBitwiseORs(int[] nums) {
        int ans = 0;
        for (int x : nums) {
            if (x % 2 == 0) {
                ans |= x;
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int evenNumberBitwiseORs(vector<int>& nums) {
        int ans = 0;
        for (int x : nums) {
            if (x % 2 == 0) {
                ans |= x;
            }
        }
        return ans;
    }
};

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func evenNumberBitwiseORs(nums []int) (ans int) {
    for _, x := range nums {
        if x%2 == 0 {
            ans |= x
        }
    }
    return
}

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function evenNumberBitwiseORs(nums: number[]): number {
    return nums.reduce((ans, x) => (x % 2 === 0 ? ans | x : ans), 0);
}

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