LeetCode #2680 — MEDIUM

Maximum OR

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy · Bit Manipulation

Example 1

[12,9]
1

Example 2

[8,1,2]
2

Step 02

Core Insight

What unlocks the optimal approach

The optimal solution should apply all the k operations on a single number.
Calculate the prefix or and the suffix or and perform k operations over each element, and maximize the answer.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2680: Maximum OR
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;
    }
}

// Accepted solution for LeetCode #2680: Maximum OR
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)
}

# Accepted solution for LeetCode #2680: Maximum OR
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

// Accepted solution for LeetCode #2680: Maximum OR
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
    }
}

// Accepted solution for LeetCode #2680: Maximum OR
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.