Bitwise OR of All Subsequence Sums

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Zomato

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ArraysBit Manipulation

You are given a 0-indexed array nums consisting of non-negative integers.

Consider the problem of calculating the sum of each subsequence of nums. A subsequence of nums is a list that can be derived from nums by deleting some or no elements without changing the order of the remaining elements. Note that the empty subsequence is considered to have a sum of 0.

Return the bitwise OR of all the subsequence sums.

Example 1:

Input: nums = [2,1,3]
Output: 7
Explanation: There are 8 possible subsequences:
- The subsequence [] has a sum of 0.
- The subsequence [2] has a sum of 2.
- The subsequence [1] has a sum of 1.
- The subsequence [3] has a sum of 3.
- The subsequence [2,1] has a sum of 3.
- The subsequence [2,3] has a sum of 5.
- The subsequence [1,3] has a sum of 4.
- The subsequence [2,1,3] has a sum of 6.
The bitwise OR of these values is 0 | 2 | 1 | 3 | 3 | 5 | 4 | 6 = 7.

Example 2:

Input: nums = [5,11,3]
Output: 15
Explanation: The bitwise OR of all subsequence sums is 15.

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 1000

Try coding it in LeetCode

Bitwise OR of All Subsequence Sums

Medium

Zomato

4 views

Topics:

ArraysBit Manipulation

You are given a 0-indexed array nums consisting of non-negative integers.

Consider the problem of calculating the sum of each subsequence of nums. A subsequence of nums is a list that can be derived from nums by deleting some or no elements without changing the order of the remaining elements. Note that the empty subsequence is considered to have a sum of 0.

Return the bitwise OR of all the subsequence sums.

Example 1:

Input: nums = [2,1,3]
Output: 7
Explanation: There are 8 possible subsequences:
- The subsequence [] has a sum of 0.
- The subsequence [2] has a sum of 2.
- The subsequence [1] has a sum of 1.
- The subsequence [3] has a sum of 3.
- The subsequence [2,1] has a sum of 3.
- The subsequence [2,3] has a sum of 5.
- The subsequence [1,3] has a sum of 4.
- The subsequence [2,1,3] has a sum of 6.
The bitwise OR of these values is 0 | 2 | 1 | 3 | 3 | 5 | 4 | 6 = 7.

Example 2:

Input: nums = [5,11,3]
Output: 15
Explanation: The bitwise OR of all subsequence sums is 15.

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 1000

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 size of the input array `nums`?
What is the maximum possible value of an element within the input array `nums`?
Can the input array `nums` be empty? If so, what should I return?
Are all numbers in the array non-negative as stated in the description, or can they be negative?
Are duplicate numbers allowed in the input array `nums`, and how would that affect the calculation of subsequence sums and their bitwise OR?

Brute Force Solution

Approach

The brute force strategy in this problem is all about trying every possible combination. We're going to find every possible subsequence sum, and then combine them in a specific way to get our final answer.

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

First, create all the possible subsequences you can make from the original set of numbers. A subsequence is any combination of numbers from the original set, including the empty set and the original set itself.
For each subsequence, calculate the sum of its numbers. This gives you a collection of subsequence sums.
Now, take all those subsequence sums and perform a bitwise OR operation on them. The bitwise OR compares the binary representations of each number. If either number has a 1 in a certain position, the result will have a 1 in that position too.
The final result of this bitwise OR operation is the answer.

Code Implementation

def bitwise_or_of_subsequence_sums(numbers):
    subsequence_sums = []

    # Generate all possible subsequences
    for i in range(1 << len(numbers)):
        subsequence = []
        current_sum = 0

        for j in range(len(numbers)):
            # Check if j-th element is included in the current subsequence
            if (i >> j) & 1:
                subsequence.append(numbers[j])
                current_sum += numbers[j]

        subsequence_sums.append(current_sum)

    # Initialize result to 0 to handle the case of an empty input
    bitwise_or_result = 0

    # Calculate bitwise OR of all subsequence sums
    # Necessary to cover all combinations
    for subsequence_sum in subsequence_sums:
        bitwise_or_result |= subsequence_sum

    return bitwise_or_result

Big(O) Analysis

Time Complexity

O(2^n) – Generating all possible subsequences from a set of n numbers results in 2^n subsequences. For each of these 2^n subsequences, we compute the sum, which takes O(n) time in the worst case (if the subsequence contains all n elements). However, the dominant operation is generating the subsequences. Finally, a bitwise OR operation is performed on all subsequence sums. Since we have 2^n subsequence sums, the OR operation takes O(2^n) time. Thus, the overall time complexity is dominated by the subsequence generation, giving us O(2^n).

Space Complexity

O(2^N) – The algorithm generates all possible subsequences of the input array. Each subsequence is a combination of elements from the original array, and there are 2^N possible subsequences for an input array of size N. To store the sums of these subsequences, the algorithm needs a data structure that can potentially hold up to 2^N integer values. Therefore, the auxiliary space complexity is O(2^N).

Optimal Solution

Approach

Instead of calculating every single possible sum of subsequences, the optimal approach cleverly uses bitwise operations to directly construct the final answer. The core idea is that if any number in the original set has a particular bit set to one, then that bit will definitely be set to one in the final result.

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

Consider each number in the given set one by one.
For each number, go through each individual bit position (like the 1's place, 2's place, 4's place, and so on).
If any number has a '1' in a particular bit position, we know that at least one of the subsequence sums must also have a '1' in that bit position.
If a '1' exists at a specific bit position in any of the numbers, guarantee that bit position will be set as one in the final answer.
Combine all such bit positions that have '1's in at least one of the numbers using the bitwise OR operation. This effectively builds the answer bit by bit.
The final result is the bitwise OR of all the numbers in the set. This single value represents the bitwise OR of all possible subsequence sums.

Code Implementation

def bitwise_or_of_subsequence_sums(numbers):
    bitwise_or_result = 0

    # Iterate through each number in the input set.
    for number in numbers:

        # Accumulate the bitwise OR. If a bit is set in any number, it will be set in result.
        bitwise_or_result |= number

    # The bitwise OR of all numbers represents bitwise OR of all subsequence sums.
    return bitwise_or_result

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through each of the n numbers in the input set once. Inside the loop, it examines the bits of each number. The number of bits examined is constant and independent of the input size (e.g., for 32-bit integers, it's 32). Because the bit examination is constant, the overall time complexity is dominated by the single loop iterating through the n numbers. Therefore, the time complexity is O(n).

Space Complexity

O(1) – The provided solution calculates the bitwise OR of all numbers in the input set directly. It does not create any auxiliary data structures like lists, hash maps, or recursion stacks. Therefore, the extra memory used is constant and independent of the input size N, where N is the number of elements in the set. The space complexity is O(1).

Edge Cases

Case	How to Handle
Empty or null input array	Return 0 since there are no subsequences and the OR of nothing is 0.
Array with a single element	Return the single element as it's the only subsequence sum.
Array with all zero values	Return 0, as all subsequence sums will be zero.
Array with large positive integers that could cause integer overflow when summed	Use a data type with sufficient capacity (e.g., long) to store intermediate sums.
Maximum sized input array (scalability)	The powerset approach inherent in subsequence generation should be optimized or a dynamic programming solution might be necessary.
Array containing one very large number and many small numbers.	The large number will likely dominate the final OR result, influencing the order of calculations.
Array containing only powers of 2.	The OR operation will result in the sum of the distinct powers of 2, potentially requiring careful analysis of bitwise operations.
Array with duplicate large values that may cause overflow if naively added repeatedly.	Using a `long` datatype will handle this overflow, and the duplication will not change the correctness of the OR operation.