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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 $nums$.

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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;
      }
  }
  

• 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;
      }
  };
  

• 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
  
  

• 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)
  }
  
  func max(a, b int) int {
  	if a > b {
  		return a
  	}
  	return b
  }
  

• 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;
  }
  
  

• 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
      }
  }
  

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