Problem Description

You are given an array nums of length n. Your task is to create a new array answer of length n - 1 where each element is calculated using the bitwise OR operation.

For each position i in the answer array (where i ranges from 0 to n - 2), you need to compute answer[i] = nums[i] | nums[i + 1]. The | symbol represents the bitwise OR operation, which compares the binary representation of two numbers bit by bit and returns 1 if at least one of the corresponding bits is 1.

For example, if nums = [1, 3, 7, 15]:

• answer[0] = 1 | 3 = 3 (binary: 01 | 11 = 11)
• answer[1] = 3 | 7 = 7 (binary: 011 | 111 = 111)
• answer[2] = 7 | 15 = 15 (binary: 0111 | 1111 = 1111)
• Result: answer = [3, 7, 15]

The solution uses Python's pairwise function to iterate through consecutive pairs of elements (a, b) in the array and applies the bitwise OR operation a | b to each pair, creating a list comprehension that generates the answer array in a single line.

Intuition

The problem asks us to perform a bitwise OR operation between every consecutive pair of elements in the array. This is a straightforward transformation where we're essentially creating a new array by combining adjacent elements.

When we think about what we need to do:

• Take element at index 0 and index 1, apply OR operation
• Take element at index 1 and index 2, apply OR operation
• Continue until we reach the last pair

This pattern of processing consecutive pairs is a common operation in programming. We need to iterate through the array but always look at two elements at a time - the current element and the next one.

The key insight is that if we have n elements in the original array, we'll have exactly n - 1 pairs of consecutive elements, which means our result array will have n - 1 elements. Each position in the result corresponds to one pair from the original array.

Python's pairwise function is perfectly suited for this task as it automatically generates consecutive pairs from an iterable. Instead of manually managing indices like nums[i] and nums[i + 1] in a loop, pairwise(nums) gives us tuples like (nums[0], nums[1]), (nums[1], nums[2]), and so on.

By combining pairwise with a list comprehension and the bitwise OR operator |, we can express the entire solution in a single, readable line that directly maps the problem statement to code.

Solution Approach

The solution uses an iteration-based approach to process consecutive pairs of elements in the array.

Implementation Steps:

1. Using the pairwise function: The solution leverages Python's pairwise function from the itertools module (available in Python 3.10+). This function takes an iterable and returns an iterator of consecutive pairs.

   • For an array [a, b, c, d], pairwise generates: (a, b), (b, c), (c, d)

2. List Comprehension: The solution uses a list comprehension to iterate through all pairs and apply the bitwise OR operation in a single expression:

   [a | b for a, b in pairwise(nums)]

3. Bitwise OR Operation: For each pair (a, b), the code calculates a | b. The bitwise OR operation works by comparing each bit position:

   • If either bit is 1, the result bit is 1
   • If both bits are 0, the result bit is 0

Algorithm Flow:

• The pairwise(nums) function automatically handles the iteration through indices 0 to n-2
• For each pair (nums[i], nums[i+1]), it computes the bitwise OR
• The list comprehension collects all results into a new list
• The resulting list has exactly n-1 elements, where n is the length of the input array

Time Complexity: O(n) where n is the length of the input array, as we iterate through the array once.

Space Complexity: O(1) extra space (not counting the output array), as we only use a constant amount of additional memory for the iteration variables.

Example Walkthrough

Let's walk through the solution with nums = [5, 2, 8, 6].

Step 1: Understanding the Input

• Array: [5, 2, 8, 6]
• Length: 4 elements
• Expected output length: 3 elements (n - 1)

Step 2: Identifying Consecutive Pairs Using pairwise(nums), we get these pairs:

• Pair 1: (5, 2) - elements at indices 0 and 1
• Pair 2: (2, 8) - elements at indices 1 and 2
• Pair 3: (8, 6) - elements at indices 2 and 3

Step 3: Applying Bitwise OR to Each Pair

For Pair 1: 5 | 2

• 5 in binary: 0101
• 2 in binary: 0010
• Bitwise OR: 0111 = 7

For Pair 2: 2 | 8

• 2 in binary: 0010
• 8 in binary: 1000
• Bitwise OR: 1010 = 10

For Pair 3: 8 | 6

• 8 in binary: 1000
• 6 in binary: 0110
• Bitwise OR: 1110 = 14

Step 4: Building the Result The list comprehension [a | b for a, b in pairwise(nums)] processes each pair:

• First iteration: 5 | 2 = 7
• Second iteration: 2 | 8 = 10
• Third iteration: 8 | 6 = 14

Final Result: [7, 10, 14]

This demonstrates how the solution efficiently transforms an array of n elements into an array of n-1 elements by applying bitwise OR to consecutive pairs.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n), where n is the length of the input array nums. The pairwise function iterates through the array once to generate consecutive pairs, and the list comprehension performs a bitwise OR operation for each pair. Since there are n-1 pairs for an array of length n, and each OR operation takes constant time O(1), the overall time complexity is O(n-1) = O(n).

Space Complexity: O(1) (excluding the output array). The pairwise function returns an iterator that generates pairs on-the-fly without creating additional storage proportional to the input size. The only extra space used is for the iterator's internal state, which is constant. If we include the output array in the space complexity analysis, it would be O(n-1) = O(n) since the result list contains n-1 elements.

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

Common Pitfalls

1. Python Version Compatibility Issue

The pairwise function is only available in Python 3.10+. If you're using an older version of Python or submitting to a platform that doesn't support Python 3.10+, the code will fail with an ImportError.

Solution: Implement your own pairwise logic using traditional iteration:

def orArray(self, nums: List[int]) -> List[int]:
    return [nums[i] | nums[i + 1] for i in range(len(nums) - 1)]

2. Empty or Single-Element Array

While the current solution handles these cases correctly (returning an empty list for arrays with length ≤ 1), developers might forget to consider edge cases when modifying the solution.

Solution: Always validate input constraints or add explicit checks if needed:

def orArray(self, nums: List[int]) -> List[int]:
    if len(nums) <= 1:
        return []
    return [a | b for a, b in pairwise(nums)]

3. Confusing Bitwise OR with Logical OR

Developers might mistakenly use logical or instead of bitwise |, which would return the first truthy value rather than performing bitwise operation.

Wrong:

return [a or b for a, b in pairwise(nums)]  # Returns first non-zero value

Correct:

return [a | b for a, b in pairwise(nums)]  # Performs bitwise OR

4. Manual Implementation with Off-by-One Errors

When implementing without pairwise, it's easy to make indexing mistakes:

Wrong:

return [nums[i] | nums[i + 1] for i in range(len(nums))]  # IndexError

Correct:

return [nums[i] | nums[i + 1] for i in range(len(nums) - 1)]  # Stops at n-2

5. Assuming Input Modification is Allowed

Some might try to modify the input array in-place to save space, but this changes the original data which may be needed elsewhere.

Avoid:

for i in range(len(nums) - 1):
    nums[i] = nums[i] | nums[i + 1]
return nums[:-1]  # Modifies original array

Better: Create a new array as shown in the original solution.