Smallest Subarrays With Maximum Bitwise OR

Medium

Amazon

24 views

Topics:

ArraysBit Manipulation

You are given a 0-indexed array nums of length n, consisting of non-negative integers. For each index i from 0 to n - 1, you must determine the size of the minimum sized non-empty subarray of nums starting at i (inclusive) that has the maximum possible bitwise OR.

In other words, let B_ij be the bitwise OR of the subarray nums[i...j]. You need to find the smallest subarray starting at i, such that bitwise OR of this subarray is equal to max(B_ik) where i <= k <= n - 1.

The bitwise OR of an array is the bitwise OR of all the numbers in it.

Return an integer array answer of size n where answer[i] is the length of the minimum sized subarray starting at i with maximum bitwise OR.

A subarray is a contiguous non-empty sequence of elements within an array.

For example:

nums = [1, 0, 2, 1, 3]

Output: [3, 3, 2, 2, 1]

Explanation: The maximum possible bitwise OR starting at any index is 3.

Starting at index 0, the shortest subarray that yields it is [1,0,2], so the length is 3.
Starting at index 1, the shortest subarray that yields the maximum bitwise OR is [0,2,1], so the length is 3.
Starting at index 2, the shortest subarray that yields the maximum bitwise OR is [2,1], so the length is 2.
Starting at index 3, the shortest subarray that yields the maximum bitwise OR is [1,3], so the length is 2.
Starting at index 4, the shortest subarray that yields the maximum bitwise OR is [3], so the length is 1.

Another example:

nums = [1, 2]

Output: [2, 1]

Could you provide an efficient solution to this problem?

Smallest Subarrays With Maximum Bitwise OR

Medium

Amazon

24 views

Topics:

ArraysBit Manipulation

You are given a 0-indexed array nums of length n, consisting of non-negative integers. For each index i from 0 to n - 1, you must determine the size of the minimum sized non-empty subarray of nums starting at i (inclusive) that has the maximum possible bitwise OR.

In other words, let B_ij be the bitwise OR of the subarray nums[i...j]. You need to find the smallest subarray starting at i, such that bitwise OR of this subarray is equal to max(B_ik) where i <= k <= n - 1.

The bitwise OR of an array is the bitwise OR of all the numbers in it.

Return an integer array answer of size n where answer[i] is the length of the minimum sized subarray starting at i with maximum bitwise OR.

A subarray is a contiguous non-empty sequence of elements within an array.

For example:

nums = [1, 0, 2, 1, 3]

Output: [3, 3, 2, 2, 1]

Explanation: The maximum possible bitwise OR starting at any index is 3.

Starting at index 0, the shortest subarray that yields it is [1,0,2], so the length is 3.
Starting at index 1, the shortest subarray that yields the maximum bitwise OR is [0,2,1], so the length is 3.
Starting at index 2, the shortest subarray that yields the maximum bitwise OR is [2,1], so the length is 2.
Starting at index 3, the shortest subarray that yields the maximum bitwise OR is [1,3], so the length is 2.
Starting at index 4, the shortest subarray that yields the maximum bitwise OR is [3], so the length is 1.

Another example:

nums = [1, 2]

Output: [2, 1]

Could you provide an efficient solution to this problem?

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum size of the input array `nums` and what is the range of values within the array?
Can the input array `nums` contain any zero values?
If there are multiple subarrays with the smallest length that achieve the maximum bitwise OR, should I return any one of them, or is there a specific requirement for which one to return?
If the array `nums` is empty, or if all elements in `nums` are zero, what value should I return?
Does the input array `nums` contain only non-negative integers?

Brute Force Solution

Approach

The brute force approach means checking every possible small group of numbers within the given list. We want to find the smallest groups that, when combined using a bitwise OR operation, give us the maximum possible value achievable from any combination of numbers in the entire list.

Here's how the algorithm would work step-by-step:

First, figure out the maximum possible OR value you can get by combining any numbers from the whole list.
Next, look at every possible small group of numbers, starting with groups of just one number.
For each small group, calculate the bitwise OR of all the numbers within that group.
If the bitwise OR of the group equals the maximum possible OR value we found earlier, then that group is a potential answer.
If the group's OR doesn't equal the maximum, throw it away and move on to the next group.
Keep track of the size of all the groups that give us the maximum OR value.
Finally, from all the groups that produce the maximum OR value, choose the smallest one. Its size is the answer we want.

Code Implementation

def smallest_subarrays_with_max_bitwise_or_brute_force(numbers):
    maximum_bitwise_or = 0
    for i in range(len(numbers)):
        current_bitwise_or = 0
        for j in range(i, len(numbers)):
            current_bitwise_or |= numbers[j]
            maximum_bitwise_or = max(maximum_bitwise_or, current_bitwise_or)

    smallest_subarray_length = float('inf')

    # Iterate through all possible subarrays
    for i in range(len(numbers)):
        for j in range(i, len(numbers)):
            current_bitwise_or = 0
            for k in range(i, j + 1):
                current_bitwise_or |= numbers[k]

            # Check if subarray's OR equals max OR
            if current_bitwise_or == maximum_bitwise_or:

                subarray_length = j - i + 1

                # Update the smallest length if needed
                smallest_subarray_length = min(smallest_subarray_length, subarray_length)

    #Handle empty array or no solution found
    if smallest_subarray_length == float('inf'):
        return 0
    return smallest_subarray_length

Big(O) Analysis

Time Complexity

O(n^3) – The algorithm first calculates the maximum OR value in O(n) time. Then, it iterates through all possible subarrays. There are approximately n^2 such subarrays (n*(n+1)/2). For each subarray, it computes the bitwise OR, which takes O(n) in the worst case (if the subarray has length n). Therefore, the overall time complexity is dominated by the subarray generation and bitwise OR calculation, resulting in O(n^2 * n) = O(n^3).

Space Complexity

O(1) – The provided brute force approach calculates the maximum OR value, then iterates through all possible subarrays. It keeps track of the minimum size of the subarrays that result in the maximum OR. The algorithm primarily uses a few variables to store the maximum OR value, the current subarray's OR value, the minimum subarray size found so far, and loop indices. These variables consume a constant amount of space regardless of the input list size (N). Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The core idea is to find the shortest possible sections that, when combined using the OR operation, give us the same result as combining all numbers together. Instead of checking every possible section, we focus only on the positions of the '1' bits in the numbers.

Here's how the algorithm would work step-by-step:

First, figure out what the final OR result should be by combining all numbers together using the OR operation. This will be our target.
Next, for each number in the input, we'll find the closest next number to the right that helps us reach the target OR value.
To do this efficiently, think about the positions where the target OR result has a '1' bit. For each number, find the furthest right number that contains each of these '1' bit positions.
The length of the shortest section needed for each number is then determined by the maximum distance to any of these rightmost positions we just found.
Return all these section lengths as the result.

Code Implementation

def smallest_subarrays_with_maximum_bitwise_or(numbers):
    target_or = 0
    for number in numbers:
        target_or |= number

    result = []
    for i in range(len(numbers)):
        rightmost_positions = []
        for bit_position in range(32):
            if (target_or >> bit_position) & 1:
                last_occurrence = -1
                for j in range(len(numbers) - 1, i - 1, -1):
                    if (numbers[j] >> bit_position) & 1:
                        last_occurrence = j
                        break
                rightmost_positions.append(last_occurrence)

        # Handle the edge case where no '1' bits are in target_or
        if not rightmost_positions:
            result.append(1)
            continue

        # Find the furthest right index.
        max_rightmost = max(rightmost_positions)

        # Determine the length of the smallest subarray.
        subarray_length = max_rightmost - i + 1

        result.append(subarray_length)

    return result

Big(O) Analysis

Time Complexity

O(n*m) – Calculating the target OR takes O(n) where n is the length of the input array. For each of the n numbers in the input array, the algorithm iterates through the 'm' positions of '1' bits in the target OR value. For each of these 'm' bits, the algorithm searches for the rightmost occurrence which can take O(n) in the worst case although usually much faster with proper data structure or early breaking. Because the search is performed m times for each of the n elements, the time complexity is O(n*m). Although the search is complex, the dominating factor is n*m.

Space Complexity

O(K) – The dominant space usage comes from storing the rightmost positions of '1' bits in the target OR value. Step 3 finds these furthest right positions and stores them. Since the target OR value's size dictates the number of '1' bits we need to track, and in the worst case, all bits could be '1', we need to store the last positions for each of these. If we denote the number of bits in the numbers in the input array as K, we are using space proportional to K to store the last positions of each '1' bit, leading to O(K) auxiliary space. The other variables used are constant in size regardless of the input array size N. Therefore, auxiliary space is O(K).

Edge Cases

Case	How to Handle
Null or empty input array	Return an empty array or throw an IllegalArgumentException, depending on the problem's specified error handling.
Input array with a single element	The smallest subarray is the element itself, so return an array containing the index of that element as a single element subarray.
All elements in the array are zero	Return an array of indices where each smallest subarray has a length of 1 and contains a zero.
All elements result in the maximum possible bitwise OR	Each element is its own smallest subarray, so return an array representing indices of individual elements.
Integer overflow in bitwise OR operation	The problem statement constrains input size to prevent integer overflow, but if it could occur, handle by using larger data types or masking.
Large input array causing performance issues	The solution uses sliding window with O(n) complexity so it can handle large input sizes, but can be optimized for cases when maximum OR is quickly reached.
Input array with duplicate values that contribute to maximum OR	The sliding window approach correctly handles duplicates, since the window expands and contracts based on satisfying the OR condition regardless of duplicate values.
No subarray exists with a bitwise OR equal to the maximum possible OR of the array.	Return an empty array or an error code if the problem specifies no such subarray exists.