Problem Description

You have an array of integers nums (0-indexed) with length n and an integer k. You can perform an operation where you select any element from the array and multiply it by 2.

Your goal is to find the maximum possible value of the bitwise OR of all elements in the array (nums[0] | nums[1] | ... | nums[n - 1]) after performing the multiplication operation at most k times.

The bitwise OR operation (|) combines bits from two numbers - if either bit is 1 at a position, the result has 1 at that position.

For example, if you have nums = [1, 2, 3] and k = 2, you could multiply one element by 2 twice (equivalent to multiplying by 4), and then calculate the OR of all elements to get your result.

The key insight is that multiplying by 2 is equivalent to a left shift operation in binary, which can significantly increase the value when combined with OR operations. You need to strategically choose which element(s) to multiply and how many times to maximize the final OR result.

Intuition

When we multiply a number by 2, we're essentially performing a left shift operation on its binary representation. For example, 5 (binary: 101) becomes 10 (binary: 1010). This introduces higher-order bits that weren't present before.

The key observation is that in bitwise OR operations, once a bit position is set to 1 by any number, it remains 1 in the final result. Therefore, to maximize the OR value, we want to create the highest possible bit positions with value 1.

Since multiplying by 2 shifts bits to the left, multiplying k times shifts bits k positions to the left (equivalent to multiplying by 2^k). To maximize our result, we should apply all k operations to the same number rather than spreading them across multiple numbers. Why? Because shifting one number k positions left creates much higher bit positions than shifting multiple numbers fewer times.

Consider this: shifting 4 (binary: 100) left by 3 positions gives 32 (binary: 100000), while shifting three different numbers by 1 position each would only double their values, creating much smaller contributions to the final OR.

Now, which number should we choose to multiply? We need to try each number as a candidate because the optimal choice depends on both the number's value and how it interacts with other numbers through the OR operation.

For each candidate number at position i, we need to compute:

• The OR of all numbers before position i (prefix OR)
• The multiplied value of nums[i] (which is nums[i] << k)
• The OR of all numbers after position i (suffix OR)

By precomputing the suffix OR values and maintaining the prefix OR as we iterate, we can efficiently check each possibility and find the maximum result.

Learn more about Greedy and Prefix Sum patterns.

Solution Approach

The implementation uses a greedy approach with preprocessing to efficiently compute the maximum OR value.

Step 1: Precompute Suffix OR Array

First, we create a suffix OR array suf where suf[i] stores the bitwise OR of all elements from index i to the end of the array. We build this array from right to left:

suf = [0] * (n + 1)
for i in range(n - 1, -1, -1):
    suf[i] = suf[i + 1] | nums[i]

Here, suf[i] = suf[i + 1] | nums[i] means the OR value at position i includes the current element nums[i] and all elements after it.

Step 2: Iterate and Calculate Maximum OR

We traverse the array from left to right while maintaining:

• pre: the prefix OR value (OR of all elements before the current position)
• ans: the maximum OR value found so far

For each position i, we calculate what the total OR would be if we applied all k operations to nums[i]:

ans = pre = 0
for i, x in enumerate(nums):
    ans = max(ans, pre | (x << k) | suf[i + 1])
    pre |= x

The expression pre | (x << k) | suf[i + 1] computes:

• pre: OR of all elements before position i
• (x << k): the value of nums[i] after multiplying by 2^k (left shift by k positions)
• suf[i + 1]: OR of all elements after position i

After checking position i, we update the prefix OR by including the current element: pre |= x.

Time and Space Complexity:

• Time: O(n) - we make two passes through the array
• Space: O(n) - for storing the suffix OR array

This approach ensures we find the optimal position to apply all k operations while efficiently computing the resulting OR value.

Example Walkthrough

Let's walk through the solution with nums = [12, 9] and k = 1.

Step 1: Build Suffix OR Array

We build the suffix array from right to left:

• suf[2] = 0 (base case, no elements)
• suf[1] = suf[2] | nums[1] = 0 | 9 = 9
• suf[0] = suf[1] | nums[0] = 9 | 12 = 13

So suf = [13, 9, 0]

To understand why 12 | 9 = 13, let's look at the binary:

• 12 in binary: 1100
• 9 in binary: 1001
• OR result: 1101 = 13

Step 2: Try Each Position for Multiplication

Initialize pre = 0 and ans = 0.

Position i = 0 (choosing to multiply 12):

• Calculate: pre | (nums[0] << k) | suf[1]
• = 0 | (12 << 1) | 9
• = 0 | 24 | 9

Let's compute 24 | 9:

• 24 in binary: 11000
• 9 in binary: 01001
• OR result: 11001 = 25

So ans = max(0, 25) = 25

Update pre = pre | nums[0] = 0 | 12 = 12

Position i = 1 (choosing to multiply 9):

• Calculate: pre | (nums[1] << k) | suf[2]
• = 12 | (9 << 1) | 0
• = 12 | 18 | 0

Let's compute 12 | 18:

• 12 in binary: 01100
• 18 in binary: 10010
• OR result: 11110 = 30

So ans = max(25, 30) = 30

Result: The maximum OR value is 30, achieved by multiplying 9 by 2 to get 18, then computing 12 | 18 = 30.

This example demonstrates why we need to try each position: multiplying the smaller number 9 actually gives us a better result than multiplying the larger number 12, because of how the bits align in the OR operation.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The algorithm consists of two main passes through the array:

• First pass: Building the suffix OR array by iterating from index n-1 to 0, which takes O(n) time
• Second pass: Iterating through the array once more to calculate the maximum OR value by considering each element as a candidate for left-shifting, which also takes O(n) time

Each operation inside the loops (bitwise OR, left shift, max comparison) takes O(1) time. Therefore, the overall time complexity is O(n) + O(n) = O(n).

Space Complexity: O(n)

The algorithm uses additional space for:

• The suffix OR array suf of size n + 1, which requires O(n) space
• A few scalar variables (ans, pre, i, x) which require O(1) space

The dominant space usage comes from the suffix array, making the overall space complexity O(n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Distributing Operations Across Multiple Elements

The Mistake: A common misconception is thinking that distributing the k operations across different elements might yield a better result. For example, with k = 4 operations, one might consider applying 2 operations each to two different elements instead of all 4 to a single element.

Why It's Wrong: The bitwise OR operation has a crucial property: once a bit position is set to 1, it remains 1 regardless of other values. Left-shifting by k positions (multiplying by 2^k) creates higher-order bits that weren't present before. Applying all operations to one element maximizes the highest bit position we can achieve.

Example:

nums = [1, 2], k = 2

# Wrong approach: Split operations
# 1 << 1 = 2, 2 << 1 = 4
# Result: 2 | 4 = 6

# Correct approach: All operations on one element
# 1 << 2 = 4, keep 2 as is
# Result: 4 | 2 = 6
# OR
# Keep 1 as is, 2 << 2 = 8
# Result: 1 | 8 = 9 (better!)

Pitfall 2: Not Considering Elements Already Having High Bits

The Mistake: Assuming the smallest element should always receive the operations, thinking it has the most "room to grow."

Why It's Wrong: An element with existing high-order bits benefits more from left-shifting because those high bits move to even higher positions. The goal is to maximize the highest bit position in the final OR result.

Example:

nums = [1, 8], k = 1

# Wrong intuition: Shift the smaller number
# 1 << 1 = 2, keep 8
# Result: 2 | 8 = 10

# Correct approach: Shift the larger number
# Keep 1, 8 << 1 = 16
# Result: 1 | 16 = 17 (better!)

Pitfall 3: Attempting to Optimize Without Prefix/Suffix Arrays

The Mistake: Trying to calculate the OR result on-the-fly for each possibility without preprocessing, leading to inefficient O(n²) solutions.

Why It's Wrong: Without the prefix and suffix OR arrays, you'd need to recalculate the OR of all elements for each position you test, resulting in unnecessary repeated calculations.

Correct Implementation Reminder: The solution correctly precomputes suffix OR values and maintains prefix OR during iteration, ensuring each position is evaluated in O(1) time after the initial O(n) preprocessing.