3173. Bitwise OR of Adjacent Elements π ο
Descriptionο
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 <= 1000 <= nums[i] <= 100
Solutionsο
Solution 1: Iterationο
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 $O(1)$.
Python3ο
class Solution:
def orArray(self, nums: List[int]) -> List[int]:
return [a | b for a, b in pairwise(nums)]
Javaο
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;
}
}
C++ο
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;
}
};
Goο
func orArray(nums []int) (ans []int) {
for i, x := range nums[1:] {
ans = append(ans, x|nums[i])
}
return
}
TypeScriptο
function orArray(nums: number[]): number[] {
return nums.slice(0, -1).map((v, i) => v | nums[i + 1]);
}