# 3173. Bitwise OR of Adjacent Elements ðŸ”’

## Description

Given an array nums of length n, return an array answer of length n - 1 such that answer[i] = nums[i] \| nums[i + 1] where \| is the bitwise OR operation.

Example 1:

Input: nums = [1,3,7,15]

Output: [3,7,15]

Example 2:

Input: nums = [8,4,2]

Output: [12,6]

Example 3:

Input: nums = [5,4,9,11]

Output: [5,13,11]

Constraints:

• 2 <= nums.length <= 100
• 0 <= nums[i] <= 100

## Solutions

### Solution 1: Iteration

We iterate through the first $n - 1$ elements of the array. For each element, we calculate the bitwise OR value of it and its next element, and store the result in the answer array.

The time complexity is $O(n)$, where $n$ is the length of the array. Ignoring the space consumption of the answer array, the space complexity is $O(1)$.

• class Solution {
public int[] orArray(int[] nums) {
int n = nums.length;
int[] ans = new int[n - 1];
for (int i = 0; i < n - 1; ++i) {
ans[i] = nums[i] | nums[i + 1];
}
return ans;
}
}

• class Solution {
public:
vector<int> orArray(vector<int>& nums) {
int n = nums.size();
vector<int> ans(n - 1);
for (int i = 0; i < n - 1; ++i) {
ans[i] = nums[i] | nums[i + 1];
}
return ans;
}
};

• class Solution:
def orArray(self, nums: List[int]) -> List[int]:
return [a | b for a, b in pairwise(nums)]


• func orArray(nums []int) (ans []int) {
for i, x := range nums[1:] {
ans = append(ans, x|nums[i])
}
return
}

• function orArray(nums: number[]): number[] {
return nums.slice(0, -1).map((v, i) => v | nums[i + 1]);
}