Find the K-or of an Array

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Bit ManipulationArrays

You are given an integer array nums, and an integer k. Let's introduce K-or operation by extending the standard bitwise OR. In K-or, a bit position in the result is set to 1 if at least k numbers in nums have a 1 in that position.

Return the K-or of nums.

Example 1:

Input: nums = [7,12,9,8,9,15], k = 4

Output: 9

Explanation:

Represent numbers in binary:

Number	Bit 3	Bit 2	Bit 1	Bit 0
7	0	1	1	1
12	1	1	0	0
9	1	0	0	1
8	1	0	0	0
9	1	0	0	1
15	1	1	1	1
Result = 9	1	0	0	1

Bit 0 is set in 7, 9, 9, and 15. Bit 3 is set in 12, 9, 8, 9, and 15.
Only bits 0 and 3 qualify. The result is (1001)₂ = 9.

Example 2:

Input: nums = [2,12,1,11,4,5], k = 6

Output: 0

Explanation: No bit appears as 1 in all six array numbers, as required for K-or with k = 6. Thus, the result is 0.

Example 3:

Input: nums = [10,8,5,9,11,6,8], k = 1

Output: 15

Explanation: Since k == 1, the 1-or of the array is equal to the bitwise OR of all its elements. Hence, the answer is 10 OR 8 OR 5 OR 9 OR 11 OR 6 OR 8 = 15.

Constraints:

1 <= nums.length <= 50
0 <= nums[i] < 2³¹
1 <= k <= nums.length

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 is the range of values for the elements in the input array, and what data type are they (e.g., integers, floats)?
What is the expected data type and range for the integer `K`?
What should the function return if the input array is empty or `K` is zero?
Are there any constraints on the size of the input array (e.g., maximum number of elements)?
Could you provide a more precise definition or example of what 'K-or' means in this context?

Brute Force Solution

Approach

The brute force approach for the K-or problem means we'll explore all possible combinations. We will analyze all possible sets of K numbers from the list, and calculate the 'or' result of each set. Finally, we will compare all these 'or' results to find the overall answer.

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

First, select the very first set of K numbers from the list.
Calculate the 'or' value of this set of K numbers.
Now, select the next possible set of K numbers.
Calculate the 'or' value of this new set.
Repeat the above two steps until you've considered every possible combination of K numbers from the list.
Compare all the 'or' values you calculated. The final answer is based on all these individual 'or' values, usually the smallest or largest, depending on the problem requirements.

Code Implementation

def find_k_or_brute_force(number_list, k_value):
    or_results = []
    number_list_length = len(number_list)

    # Iterate through all possible combinations of k_value elements
    for i in range(1 << number_list_length):
        current_combination = []
        for j in range(number_list_length):
            if (i >> j) & 1:
                current_combination.append(number_list[j])

        if len(current_combination) == k_value:
            # We have a valid combination of size k_value
            current_or = 0

            for number in current_combination:
                current_or |= number

            or_results.append(current_or)

    # Determine the final result based on the 'or' values.
    if not or_results:
        return 0

    return min(or_results)

Big(O) Analysis

Time Complexity

O(n^k) – The brute force approach involves iterating through all possible combinations of K elements from an array of size n. Calculating the number of such combinations involves calculating "n choose k", which equals n! / (k! * (n-k)!). The outer loop has to choose K items. Within the loop calculating the 'or' of the chosen K items takes O(K) time. Since the dominating factor is choosing all the sets of K items from n, the time complexity is primarily determined by the combinatorial factor, which is O(n^k) because we are selecting K items and the number of combinations is affected exponentially.

Space Complexity

O(1) – The provided brute force approach calculates the 'or' value for each combination of K numbers. While it iterates through different combinations, it doesn't appear to store all of these 'or' values simultaneously. It likely keeps track of one single final 'or' value (smallest or largest) and updates it as it iterates. No auxiliary data structures are explicitly mentioned or implied to store intermediate sets or 'or' values corresponding to the size of the input array N, or K. Therefore, the auxiliary space used is constant.

Optimal Solution

Approach

The key idea is to analyze each bit position separately. We can significantly reduce computation by focusing on bits that are guaranteed to contribute to the final result and ignoring the others.

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

Imagine the numbers as made up of individual bits, like building blocks.
Start examining the bits from the least significant (rightmost) to the most significant (leftmost).
For each bit position, count how many numbers in the array have a '1' in that position.
If the count is greater than or equal to the given number 'K', this means that bit position will be a '1' in our final answer. Because at least K numbers have a 1 at that position, the 'or' operation will set that bit to 1.
If the count is less than 'K', that bit position will be a '0' in the final answer.
Continue this process for each bit position.
Combine the bits we determined to be '1' and '0' to create the final number, which is the K-or of the array.

Code Implementation

def find_k_or(numbers, k_value):
    result = 0
    max_bit_length = 0

    for number in numbers:
        bit_length = number.bit_length()
        if bit_length > max_bit_length:
            max_bit_length = bit_length

    for bit_position in range(max_bit_length): 
        count = 0
        for number in numbers:
            # Check if the bit is set at current position
            if (number >> bit_position) & 1:
                count += 1

        # If the count is >= k_value, set the bit in the result
        if count >= k_value:
            result |= (1 << bit_position)

    return result

Big(O) Analysis

Time Complexity

O(n * log(max_value)) – The algorithm iterates through each bit position of the numbers in the input array. The outer loop effectively runs from the least significant bit to the most significant bit of the largest number in the array, giving us a loop that runs log(max_value) times, where max_value is the largest number in the array. Inside this loop, we iterate through all 'n' numbers in the array to count the number of '1's at the current bit position. Therefore, the total time complexity is O(n * log(max_value)).

Space Complexity

O(1) – The described algorithm iterates through the bit positions of the numbers in the input array. It uses a counter variable to keep track of the number of ones at each bit position and accumulates the K-or result in another variable. These variables require constant extra space regardless of the size of the input array (N). Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty array	Return 0, as there are no elements to perform bitwise OR on.
k is zero	The K-or will always be 0 in this case, so return 0.
k is larger than the array length	Return 0, as there will be less than k elements and the condition cannot be met.
Array contains negative numbers	The bitwise OR operation will work correctly with negative numbers represented in two's complement.
Array contains only zeros	The K-or will be 0 since the bitwise OR of any number of zeros is zero.
Large array size causing performance issues	Optimize the algorithm to have a time complexity of O(n log n) or better to handle large input sizes efficiently, possibly using sorting or priority queues.
Integer overflow during bitwise OR operation	Use a data type that can hold the result of the bitwise OR operation without overflowing (e.g., long long in C++, long in Java).
All elements in the array are the same	The K-or will be the bitwise OR of k instances of the same number.

Null or empty array

How to Handle:

Return 0, as there are no elements to perform bitwise OR on.

k is zero

How to Handle:

The K-or will always be 0 in this case, so return 0.

k is larger than the array length

How to Handle:

Return 0, as there will be less than k elements and the condition cannot be met.

Array contains negative numbers

How to Handle:

The bitwise OR operation will work correctly with negative numbers represented in two's complement.

Array contains only zeros

How to Handle:

The K-or will be 0 since the bitwise OR of any number of zeros is zero.

Large array size causing performance issues

How to Handle:

Optimize the algorithm to have a time complexity of O(n log n) or better to handle large input sizes efficiently, possibly using sorting or priority queues.

Integer overflow during bitwise OR operation

How to Handle:

Use a data type that can hold the result of the bitwise OR operation without overflowing (e.g., long long in C++, long in Java).

All elements in the array are the same

How to Handle:

The K-or will be the bitwise OR of k instances of the same number.