README_EN.md

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3173.Bitwise%20OR%20of%20Adjacent%20Elements/README_EN.md	Bit Manipulation Array

3173. Bitwise OR of Adjacent Elements 🔒

中文文档

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 $O(1)$.

Python3

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

Java

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++

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;
    }
};

Go

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

TypeScript

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