1class Solution {
2    /**
3     * Computes the bitwise OR of all even numbers in the given array.
4     *
5     * @param nums the input array of integers
6     * @return the bitwise OR of all even elements; 0 if no even element exists
7     */
8    public int evenNumberBitwiseORs(int[] nums) {
9        // Accumulator for the bitwise OR result, initialized to 0
10        // (0 is the identity element for the OR operation)
11        int result = 0;
12
13        // Iterate over each number in the array
14        for (int num : nums) {
15            // Check whether the current number is even
16            if (num % 2 == 0) {
17                // Merge the even number into the result using bitwise OR
18                result |= num;
19            }
20        }
21
22        // Return the accumulated bitwise OR of all even numbers
23        return result;
24    }
25}
26

1class Solution {
2public:
3    int evenNumberBitwiseORs(vector<int>& nums) {
4        // Initialize the result to 0, which is the identity element for bitwise OR
5        int result = 0;
6
7        // Iterate over each number in the input array
8        for (int num : nums) {
9            // Check whether the current number is even
10            if (num % 2 == 0) {
11                // Accumulate the even number into the result using bitwise OR
12                result |= num;
13            }
14        }
15
16        // Return the bitwise OR of all even numbers in the array
17        return result;
18    }
19};
20

1/**
2 * Computes the bitwise OR of all even numbers in the given array.
3 *
4 * Approach:
5 * 1. Iterate through the array using `reduce`, starting with an accumulator of 0.
6 * 2. For each element, check whether it is even (divisible by 2).
7 * 3. If even, merge it into the accumulator using the bitwise OR operator.
8 * 4. If odd, skip it and keep the accumulator unchanged.
9 *
10 * Time Complexity:  O(n), where n is the length of the array (single pass).
11 * Space Complexity: O(1), only a constant amount of extra space is used.
12 *
13 * @param nums - The input array of integers.
14 * @returns The bitwise OR of all even numbers; returns 0 if no even number exists.
15 */
16function evenNumberBitwiseORs(nums: number[]): number {
17    return nums.reduce(
18        (ans: number, num: number): number =>
19            // Include `num` in the result only when it is even
20            num % 2 === 0 ? ans | num : ans,
21        0 // Initial accumulator value (identity element for bitwise OR)
22    );
23}
24```
25
26**Explanation of the changes:**
27
281. **Naming standardization**: The loop variable `x` was renamed to `num`, which more clearly conveys that it represents a numeric element from the input array. The function name `evenNumberBitwiseORs` is preserved as required.
29
302. **Explicit type annotations**: Added types to the `reduce` callback parameters and return value (`ans: number, num: number): number`) to make full use of TypeScript's type system.
31
323. **Comments added**: A JSDoc block documents the purpose, approach, complexity, parameters, and return value, while inline comments clarify the even-number check and the initial accumulator.
33
34**Alternative perspective** — if you prefer an imperative style, the same logic can be written with a simple loop, which some find easier to read and debug:
35
36```typescript
37function evenNumberBitwiseORs(nums: number[]): number {
38    let ans = 0;
39    for (const num of nums) {
40        if (num % 2 === 0) {
41            ans |= num; // Accumulate even numbers via bitwise OR
42        }
43    }
44    return ans;
45}
46