2680. Maximum OR
Description
You are given a 0-indexed integer array nums
of length n
and an integer k
. In an operation, you can choose an element and multiply it by 2
.
Return the maximum possible value of nums[0] | nums[1] | ... | nums[n - 1]
that can be obtained after applying the operation on nums at most k
times.
Note that a | b
denotes the bitwise or between two integers a
and b
.
Example 1:
Input: nums = [12,9], k = 1 Output: 30 Explanation: If we apply the operation to index 1, our new array nums will be equal to [12,18]. Thus, we return the bitwise or of 12 and 18, which is 30.
Example 2:
Input: nums = [8,1,2], k = 2 Output: 35 Explanation: If we apply the operation twice on index 0, we yield a new array of [32,1,2]. Thus, we return 32|1|2 = 35.
Constraints:
1 <= nums.length <= 105
1 <= nums[i] <= 109
1 <= k <= 15
Solutions
Solution 1: Greedy + Preprocessing
We notice that in order to maximize the answer, we should apply $k$ times of bitwise OR to the same number.
First, we preprocess the suffix OR value array $suf$ of the array $nums$, where $suf[i]$ represents the bitwise OR value of $nums[i], nums[i + 1], \cdots, nums[n - 1]$.
Next, we traverse the array $nums$ from left to right, and maintain the current prefix OR value $pre$. For the current position $i$, we perform $k$ times of bitwise left shift on $nums[i]$, i.e., $nums[i] \times 2^k$, and perform bitwise OR operation with $pre$ to obtain the intermediate result. Then, we perform bitwise OR operation with $suf[i + 1]$ to obtain the maximum OR value with $nums[i]$ as the last number. By enumerating all possible positions $i$, we can obtain the final answer.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.
1 2 3 4 5 6 7 8 9 10 11 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 |
|