LeetCode 2680. Maximum OR Solution in Java, C++, Python & More | Explanation + Code

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], …, 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] × 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.

PythonJavaC++GoTypeScriptRust

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(code-box)

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; } }(code-box)

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; } };(code-box)

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) }(code-box)

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; }(code-box)

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 } }(code-box)