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Find Subarray With Bitwise OR Closest to K

#23 Most AskedHard

Find Subarray With Bitwise OR Closest to K #23 Most Asked

Hard

Salesforce

4 views

Topics:

ArraysBit Manipulation

You are given an array nums and an integer k. You need to find a subarray of nums such that the absolute difference between k and the bitwise OR of the subarray elements is as small as possible. In other words, select a subarray nums[l..r] such that |k - (nums[l] OR nums[l + 1] ... OR nums[r])| is minimum.

Return the minimum possible value of the absolute difference.

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

Example 1:

Input: nums = [1,2,4,5], k = 3

Output: 0

Explanation:

The subarray nums[0..1] has OR value 3, which gives the minimum absolute difference |3 - 3| = 0.

Example 2:

Input: nums = [1,3,1,3], k = 2

Output: 1

Explanation:

The subarray nums[1..1] has OR value 3, which gives the minimum absolute difference |3 - 2| = 1.

Example 3:

Input: nums = [1], k = 10

Output: 9

Explanation:

There is a single subarray with OR value 1, which gives the minimum absolute difference |10 - 1| = 9.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹
1 <= k <= 10⁹

Try coding it in LeetCode

Find Subarray With Bitwise OR Closest to K

#23 Most AskedHard

Find Subarray With Bitwise OR Closest to K #23 Most Asked

Hard

Salesforce

4 views

Topics:

ArraysBit Manipulation

Return the minimum possible value of the absolute difference.

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

Example 1:

Input: nums = [1,2,4,5], k = 3

Output: 0

Explanation:

The subarray nums[0..1] has OR value 3, which gives the minimum absolute difference |3 - 3| = 0.

Example 2:

Input: nums = [1,3,1,3], k = 2

Output: 1

Explanation:

The subarray nums[1..1] has OR value 3, which gives the minimum absolute difference |3 - 2| = 1.

Example 3:

Input: nums = [1], k = 10

Output: 9

Explanation:

There is a single subarray with OR value 1, which gives the minimum absolute difference |10 - 1| = 9.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹
1 <= k <= 10⁹

Try coding it in LeetCode

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 are the constraints on the values within the `nums` array and the value of `k`? Specifically, what are the upper and lower bounds?
If multiple subarrays have the same minimum absolute difference between their bitwise OR and `k`, should I return the shortest one? What should I return if multiple shortest subarrays exist with the same minimal difference?
What should I return if the input array `nums` is empty?
Can the target integer `k` be zero?
Does the problem require finding the *absolute* closest value, or are smaller bitwise OR values preferred if two subarrays are equidistant from `k`?

Brute Force Solution

Approach

The brute force approach to this problem involves checking every possible group of consecutive numbers within the given list. For each group, we'll calculate a special value using the bitwise OR operation and compare that value to the target number.

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

First, consider every possible group starting from the beginning of the list. This means looking at the first number alone, then the first two numbers together, then the first three, and so on, until you've included all the numbers in the list.
For each of these groups, calculate the bitwise OR value. Think of it like combining the numbers in a special way using their binary representations.
Compare this calculated value to the target number and note how close it is. The closer, the better.
Now, do the same thing starting from the second number in the list. Consider the second number alone, then the second and third numbers together, then the second, third, and fourth, and so on.
Continue this process, moving further down the list, until you've considered all possible groups of consecutive numbers.
Keep track of the group whose bitwise OR value is closest to the target number.
After examining all possible groups, select the group that resulted in the bitwise OR value closest to the target. That's your answer.

Code Implementation

def find_closest_subarray_brute_force(numbers, target):    closest_bitwise_or_value = float('inf')
    closest_subarray = []
    
    # Iterate through all possible start indices
    for start_index in range(len(numbers)):
        current_bitwise_or = 0
        
        # Iterate through all possible end indices starting from start_index
        for end_index in range(start_index, len(numbers)):
            # Calculate the bitwise OR of the current subarray
            current_bitwise_or |= numbers[end_index]

            # Check if the current bitwise OR is closer to the target            if abs(current_bitwise_or - target) < abs(closest_bitwise_or_value - target):
                closest_bitwise_or_value = current_bitwise_or

    return closest_bitwise_or_value

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through all possible subarrays of the input array of size n. The outer loop selects the starting index of the subarray, which runs n times. The inner loop extends the subarray from the starting index to the end of the array, which on average runs n/2 times. Thus, the total number of iterations is approximately n * (n/2), which simplifies to O(n²). The bitwise OR operation within the inner loop takes constant time and does not affect the overall time complexity.

Space Complexity

O(1) – The brute force approach described only uses a few variables to store the current bitwise OR value, the minimum difference found so far, and potentially the start and end indices of the best subarray. These variables take up constant space, irrespective of the input array's size (N). Therefore, the algorithm's auxiliary space complexity is O(1).

Optimal Solution

Approach

The key is to realize that the bitwise OR of a subarray will only ever increase as you add more elements. We can leverage this property to efficiently find the subarray whose bitwise OR is closest to the target value without checking every possible subarray. We use a sliding window approach to achieve this.

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

Start by creating a 'window' that represents a potential subarray.
Move the 'end' of the window to the right, one element at a time. Calculate the bitwise OR of all numbers within the current window.
If the bitwise OR is less than the target value, keep expanding the window.
If the bitwise OR becomes greater than the target value, start shrinking the window from the left by moving the 'start' of the window to the right until the bitwise OR is no longer greater than the target.
While the window is being adjusted, keep track of the bitwise OR that is closest to the target value, and the boundaries of that window.
Continue moving the 'end' of the window through the input, expanding and shrinking as needed, and updating the closest bitwise OR found so far.
After processing the entire input, you'll have found the subarray whose bitwise OR is closest to the target value.

Code Implementation

def find_closest_bitwise_or_subarray(numbers, target):
    closest_subarray = []
    closest_bitwise_or = float('inf')
    min_difference = float('inf')

    window_start = 0
    current_bitwise_or = 0

    for window_end in range(len(numbers)):
        current_bitwise_or |= numbers[window_end]

        # Shrink the window if the current bitwise OR exceeds the target
        while current_bitwise_or > target and window_start <= window_end:

            #This ensures we get closer to the target
            current_bitwise_or ^= numbers[window_start]
            window_start += 1

        difference = abs(current_bitwise_or - target)

        # Update result if current subarray is closer to target
        if difference < min_difference:
            min_difference = difference
            closest_bitwise_or = current_bitwise_or
            closest_subarray = numbers[window_start:window_end+1]
        elif difference == min_difference and len(numbers[window_start:window_end+1]) < len(closest_subarray):
            #If there is a tie, return the smallest subarray
            closest_bitwise_or = current_bitwise_or
            closest_subarray = numbers[window_start:window_end+1]

    # If the numbers array is empty
    if not numbers:
        return []

    return closest_subarray

Big(O) Analysis

Time Complexity

O(n) – The solution employs a sliding window approach. The 'end' of the window iterates through the array once (n steps). While the 'end' pointer always moves forward, the 'start' pointer might move forward as well to shrink the window. Importantly, the 'start' pointer can only catch up to the 'end' pointer, ensuring that it also moves forward at most n steps in total. Thus, both pointers move at most n steps, resulting in O(n) time complexity.

Space Complexity

O(1) – The algorithm uses a sliding window approach with a fixed number of variables: start, end, current bitwise OR, closest bitwise OR, and the boundaries of the closest window. The space required for these variables does not depend on the input array size N. Therefore, the auxiliary space complexity is constant.

Edge Cases

Case	How to Handle
Empty input array	Return 0 as the length of the shortest subarray if the input array is empty since there is no subarray.
Target k is 0	The algorithm should correctly identify subarrays with a bitwise OR value closest to 0, prioritizing shorter subarrays.
Array contains only zeros	If k is also 0, return the length of the smallest subarray (1); otherwise, the closest OR will be 0.
Array contains very large numbers (approaching integer limits)	Ensure that bitwise OR operations don't cause integer overflow, potentially by using larger data types if necessary, or limiting the length of the subarray considered.
Target k is a very large number (approaching integer limit)	The algorithm should handle target values that are near the maximum possible integer value without overflow or incorrect calculations.
All numbers in the array are the same	The closest OR will either be the number itself or 0, depending on the value of k; handle this uniformly.
Array contains a single element	Compare the single element with k and return 1, the length of this subarray.
Multiple subarrays have the same minimum difference from k, differing in length.	The algorithm needs to keep track of both the minimum difference seen so far and the shortest subarray length that achieves it, updating the length only when a strictly smaller difference is found or an equal difference with a shorter length.

Empty input array

How to Handle:

Return 0 as the length of the shortest subarray if the input array is empty since there is no subarray.

Target k is 0

How to Handle:

The algorithm should correctly identify subarrays with a bitwise OR value closest to 0, prioritizing shorter subarrays.

Array contains only zeros

How to Handle:

If k is also 0, return the length of the smallest subarray (1); otherwise, the closest OR will be 0.

Array contains very large numbers (approaching integer limits)

How to Handle:

Ensure that bitwise OR operations don't cause integer overflow, potentially by using larger data types if necessary, or limiting the length of the subarray considered.

Target k is a very large number (approaching integer limit)

How to Handle:

The algorithm should handle target values that are near the maximum possible integer value without overflow or incorrect calculations.

All numbers in the array are the same

How to Handle:

The closest OR will either be the number itself or 0, depending on the value of k; handle this uniformly.

Array contains a single element

How to Handle:

Compare the single element with k and return 1, the length of this subarray.

Multiple subarrays have the same minimum difference from k, differing in length.

How to Handle:

The algorithm needs to keep track of both the minimum difference seen so far and the shortest subarray length that achieves it, updating the length only when a strictly smaller difference is found or an equal difference with a shorter length.

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 are the constraints on the values within the `nums` array and the value of `k`? Specifically, what are the upper and lower bounds?
If multiple subarrays have the same minimum absolute difference between their bitwise OR and `k`, should I return the shortest one? What should I return if multiple shortest subarrays exist with the same minimal difference?
What should I return if the input array `nums` is empty?
Can the target integer `k` be zero?
Does the problem require finding the *absolute* closest value, or are smaller bitwise OR values preferred if two subarrays are equidistant from `k`?

Brute Force Solution

Approach

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

First, consider every possible group starting from the beginning of the list. This means looking at the first number alone, then the first two numbers together, then the first three, and so on, until you've included all the numbers in the list.
For each of these groups, calculate the bitwise OR value. Think of it like combining the numbers in a special way using their binary representations.
Compare this calculated value to the target number and note how close it is. The closer, the better.
Now, do the same thing starting from the second number in the list. Consider the second number alone, then the second and third numbers together, then the second, third, and fourth, and so on.
Continue this process, moving further down the list, until you've considered all possible groups of consecutive numbers.
Keep track of the group whose bitwise OR value is closest to the target number.
After examining all possible groups, select the group that resulted in the bitwise OR value closest to the target. That's your answer.

Code Implementation

def find_closest_subarray_brute_force(numbers, target):    closest_bitwise_or_value = float('inf')
    closest_subarray = []
    
    # Iterate through all possible start indices
    for start_index in range(len(numbers)):
        current_bitwise_or = 0
        
        # Iterate through all possible end indices starting from start_index
        for end_index in range(start_index, len(numbers)):
            # Calculate the bitwise OR of the current subarray
            current_bitwise_or |= numbers[end_index]

            # Check if the current bitwise OR is closer to the target            if abs(current_bitwise_or - target) < abs(closest_bitwise_or_value - target):
                closest_bitwise_or_value = current_bitwise_or

    return closest_bitwise_or_value

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Start by creating a 'window' that represents a potential subarray.
Move the 'end' of the window to the right, one element at a time. Calculate the bitwise OR of all numbers within the current window.
If the bitwise OR is less than the target value, keep expanding the window.
If the bitwise OR becomes greater than the target value, start shrinking the window from the left by moving the 'start' of the window to the right until the bitwise OR is no longer greater than the target.
While the window is being adjusted, keep track of the bitwise OR that is closest to the target value, and the boundaries of that window.
Continue moving the 'end' of the window through the input, expanding and shrinking as needed, and updating the closest bitwise OR found so far.
After processing the entire input, you'll have found the subarray whose bitwise OR is closest to the target value.

Code Implementation

def find_closest_bitwise_or_subarray(numbers, target):
    closest_subarray = []
    closest_bitwise_or = float('inf')
    min_difference = float('inf')

    window_start = 0
    current_bitwise_or = 0

    for window_end in range(len(numbers)):
        current_bitwise_or |= numbers[window_end]

        # Shrink the window if the current bitwise OR exceeds the target
        while current_bitwise_or > target and window_start <= window_end:

            #This ensures we get closer to the target
            current_bitwise_or ^= numbers[window_start]
            window_start += 1

        difference = abs(current_bitwise_or - target)

        # Update result if current subarray is closer to target
        if difference < min_difference:
            min_difference = difference
            closest_bitwise_or = current_bitwise_or
            closest_subarray = numbers[window_start:window_end+1]
        elif difference == min_difference and len(numbers[window_start:window_end+1]) < len(closest_subarray):
            #If there is a tie, return the smallest subarray
            closest_bitwise_or = current_bitwise_or
            closest_subarray = numbers[window_start:window_end+1]

    # If the numbers array is empty
    if not numbers:
        return []

    return closest_subarray

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array	Return 0 as the length of the shortest subarray if the input array is empty since there is no subarray.
Target k is 0	The algorithm should correctly identify subarrays with a bitwise OR value closest to 0, prioritizing shorter subarrays.
Array contains only zeros	If k is also 0, return the length of the smallest subarray (1); otherwise, the closest OR will be 0.
Array contains very large numbers (approaching integer limits)	Ensure that bitwise OR operations don't cause integer overflow, potentially by using larger data types if necessary, or limiting the length of the subarray considered.
Target k is a very large number (approaching integer limit)	The algorithm should handle target values that are near the maximum possible integer value without overflow or incorrect calculations.
All numbers in the array are the same	The closest OR will either be the number itself or 0, depending on the value of k; handle this uniformly.
Array contains a single element	Compare the single element with k and return 1, the length of this subarray.
Multiple subarrays have the same minimum difference from k, differing in length.	The algorithm needs to keep track of both the minimum difference seen so far and the shortest subarray length that achieves it, updating the length only when a strictly smaller difference is found or an equal difference with a shorter length.

Empty input array

How to Handle:

Return 0 as the length of the shortest subarray if the input array is empty since there is no subarray.

Target k is 0

How to Handle:

The algorithm should correctly identify subarrays with a bitwise OR value closest to 0, prioritizing shorter subarrays.

Array contains only zeros

How to Handle:

If k is also 0, return the length of the smallest subarray (1); otherwise, the closest OR will be 0.

Array contains very large numbers (approaching integer limits)

How to Handle:

Ensure that bitwise OR operations don't cause integer overflow, potentially by using larger data types if necessary, or limiting the length of the subarray considered.

Target k is a very large number (approaching integer limit)

How to Handle:

The algorithm should handle target values that are near the maximum possible integer value without overflow or incorrect calculations.

All numbers in the array are the same

How to Handle:

The closest OR will either be the number itself or 0, depending on the value of k; handle this uniformly.

Array contains a single element

How to Handle:

Compare the single element with k and return 1, the length of this subarray.

Multiple subarrays have the same minimum difference from k, differing in length.

How to Handle:

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