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 <= 1051 <= nums[i] <= 1091 <= 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 $nums$.
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| class Solution:
def maximumOr(self, nums: List[int], k: int) -> int:
n = len(nums)
suf = [0] * (n + 1)
for i in range(n - 1, -1, -1):
suf[i] = suf[i + 1] | nums[i]
ans = pre = 0
for i, x in enumerate(nums):
ans = max(ans, pre | (x << k) | suf[i + 1])
pre |= x
return ans
|
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| class Solution {
public long maximumOr(int[] nums, int k) {
int n = nums.length;
long[] suf = new long[n + 1];
for (int i = n - 1; i >= 0; --i) {
suf[i] = suf[i + 1] | nums[i];
}
long ans = 0, pre = 0;
for (int i = 0; i < n; ++i) {
ans = Math.max(ans, pre | (1L * nums[i] << k) | suf[i + 1]);
pre |= nums[i];
}
return ans;
}
}
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| class Solution {
public:
long long maximumOr(vector<int>& nums, int k) {
int n = nums.size();
long long suf[n + 1];
memset(suf, 0, sizeof(suf));
for (int i = n - 1; i >= 0; --i) {
suf[i] = suf[i + 1] | nums[i];
}
long long ans = 0, pre = 0;
for (int i = 0; i < n; ++i) {
ans = max(ans, pre | (1LL * nums[i] << k) | suf[i + 1]);
pre |= nums[i];
}
return ans;
}
};
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| func maximumOr(nums []int, k int) int64 {
n := len(nums)
suf := make([]int, n+1)
for i := n - 1; i >= 0; i-- {
suf[i] = suf[i+1] | nums[i]
}
ans, pre := 0, 0
for i, x := range nums {
ans = max(ans, pre|(nums[i]<<k)|suf[i+1])
pre |= x
}
return int64(ans)
}
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| function maximumOr(nums: number[], k: number): number {
const n = nums.length;
const suf: bigint[] = Array(n + 1).fill(0n);
for (let i = n - 1; i >= 0; i--) {
suf[i] = suf[i + 1] | BigInt(nums[i]);
}
let [ans, pre] = [0, 0n];
for (let i = 0; i < n; i++) {
ans = Math.max(Number(ans), Number(pre | (BigInt(nums[i]) << BigInt(k)) | suf[i + 1]));
pre |= BigInt(nums[i]);
}
return ans;
}
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| impl Solution {
pub fn maximum_or(nums: Vec<i32>, k: i32) -> i64 {
let n = nums.len();
let mut suf = vec![0; n + 1];
for i in (0..n).rev() {
suf[i] = suf[i + 1] | (nums[i] as i64);
}
let mut ans = 0i64;
let mut pre = 0i64;
let k64 = k as i64;
for i in 0..n {
ans = ans.max(pre | ((nums[i] as i64) << k64) | suf[i + 1]);
pre |= nums[i] as i64;
}
ans
}
}
|
