#2680

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Maximum OR

Maximize the bitwise OR of an array by applying at most k multiplication operations on selected elements.

array greedy bit manipulation prefix sum

Problem Statement

You are given a 0-indexed integer array nums of length n and an integer k. You can apply at most k operations on the elements of nums, where each operation involves multiplying one element of nums by 2.

The goal is to maximize the bitwise OR of all the elements of nums after performing the operations. You must return the maximum OR value achievable after at most k operations.

Examples

Example 1

Input: nums = [12,9], k = 1

Output: 30

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

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

Solution Approach

Greedy Approach with Focus on One Element

The optimal solution applies all k operations to a single element of the array. By focusing on one element, we maximize the power of that element's value, which contributes most significantly to the OR result.

Bitwise OR Understanding

Bitwise OR results in a value that contains all the bits set in either of the operands. A greedy approach will ensure that the element receiving the most operations contributes the highest possible value, affecting the final OR outcome.

Maximizing with Prefix Sum

By calculating a prefix sum of the OR values, we can efficiently track and update the OR result as we modify elements. This helps in reducing recalculations and ensures we are optimizing the operation usage.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the approach chosen for calculating the OR value and applying operations. A naive approach could be O(n * k), but optimized solutions may reduce this to O(n log(max(nums))) with careful manipulation of the OR and operations.

What Interviewers Usually Probe

Candidate should demonstrate understanding of bitwise operations and greedy algorithms.
Look for efficient handling of the k operations on a single element to maximize OR.
Candidate may struggle if they fail to prioritize which element to operate on.

Common Pitfalls or Variants

Common pitfalls

Failing to apply all k operations on a single element, leading to suboptimal OR values.
Not recognizing the importance of bitwise OR and how each operation doubles the value of an element.
Inefficient brute-force approaches that involve recalculating the OR too often.

Follow-up variants

Allowing operations on multiple elements, not just one.
Changing the type of operation (e.g., multiplying by a factor other than 2).
Considering a scenario where the OR operation also involves other bit manipulations (e.g., XOR).

FAQ

How does the greedy approach work in the Maximum OR problem?

The greedy approach applies all k operations to a single element to maximize its value, which contributes most to the final OR result.

What is the optimal solution for the Maximum OR problem?

The optimal solution focuses all k operations on a single element, maximizing its value and subsequently the bitwise OR result.

Can the Maximum OR problem be solved without bitwise operations?

No, bitwise OR is central to the problem, as it determines the final value after applying operations on the elements.

What is the time complexity of the Maximum OR problem?

The time complexity can range from O(n * k) to O(n log(max(nums))) depending on the approach and optimizations applied.

How does GhostInterview help with the Maximum OR problem?

GhostInterview provides insights into the greedy approach and helps refine time complexity analysis, offering structured hints and solutions.

terminal

Solution

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.

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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

Continue Practicing

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