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XOR Operation in an Array #19 Most Asked

Easy

Walmart

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Topics:

ArraysBit Manipulation

You are given an integer n and an integer start.

Define an array nums where nums[i] = start + 2 * i (0-indexed) and n == nums.length.

Return the bitwise XOR of all elements of nums.

Example 1:

Input: n = 5, start = 0
Output: 8
Explanation: Array nums is equal to [0, 2, 4, 6, 8] where (0 ^ 2 ^ 4 ^ 6 ^ 8) = 8.
Where "^" corresponds to bitwise XOR operator.

Example 2:

Input: n = 4, start = 3
Output: 8
Explanation: Array nums is equal to [3, 5, 7, 9] where (3 ^ 5 ^ 7 ^ 9) = 8.

Constraints:

1 <= n <= 1000
0 <= start <= 1000
n == nums.length

Try coding it in LeetCode

XOR Operation in an Array #19 Most Asked

Easy

Walmart

3 views

Topics:

ArraysBit Manipulation

You are given an integer n and an integer start.

Define an array nums where nums[i] = start + 2 * i (0-indexed) and n == nums.length.

Return the bitwise XOR of all elements of nums.

Example 1:

Input: n = 5, start = 0
Output: 8
Explanation: Array nums is equal to [0, 2, 4, 6, 8] where (0 ^ 2 ^ 4 ^ 6 ^ 8) = 8.
Where "^" corresponds to bitwise XOR operator.

Example 2:

Input: n = 4, start = 3
Output: 8
Explanation: Array nums is equal to [3, 5, 7, 9] where (3 ^ 5 ^ 7 ^ 9) = 8.

Constraints:

1 <= n <= 1000
0 <= start <= 1000
n == 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 are the constraints for `n` and `start`? Specifically, what are the minimum and maximum possible values for each?
Can `n` be zero? If so, what should the function return?
Is `start` guaranteed to be an integer, or could it be a floating-point number?
Could you provide a few example inputs and their corresponding expected outputs to confirm my understanding of the problem?
Are there any memory constraints I should be aware of given the potential size of `n`?

Brute Force Solution

Approach

The brute force method means we explore every possible outcome. In this case, we calculate the result of the XOR operation for all possible starting points and lengths within the given data. We then compare these results to find the one we're looking for.

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

Consider each element in the collection as a potential starting point.
From that starting point, calculate the XOR result by combining it with the next element, then with the next-next element, and so on until you reach the end of the collection.
Store the XOR result you calculated from each starting point and length.
Compare all the stored XOR results to see if any of them match the target value we're seeking.
If a match is found, that is our answer. If you exhaust all possibilities and no match is found, then no such result exists within the collection.

Code Implementation

def find_xor_brute_force(numbers, target):
    # Initialize the running total with zero as described in the instructions
    running_total = 0

    def explore_combinations(index, current_total):
        nonlocal running_total

        # Base case: If we reach the end of the array, check if target is found.
        if index == len(numbers):
            if current_total == target:
                return True
            else:
                return False

        # Explore including the current number in the XOR operation.
        if explore_combinations(index + 1, current_total ^ numbers[index]):
            return True

        # Explore excluding the current number in the XOR operation.
        # We need to explore every possible combination.
        if explore_combinations(index + 1, current_total):
            return True

        return False

    # Start exploring combinations from the beginning of the array.
    if explore_combinations(0, running_total):
        return True
    else:
        return False

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each of the n elements in the array as a potential starting point for the XOR operation. For each starting element, it then iterates through the remaining elements to compute the XOR sum of all possible subarrays starting from that element. In the worst case, for the first element, it computes XORs for n-1 subarrays, for the second n-2, and so on. This results in approximately n * (n-1) / 2 operations, which simplifies to O(n²).

Space Complexity

O(N^2) – The brute force algorithm stores XOR results calculated from each starting point and length. Since we iterate through each element as a potential starting point, and from each starting point, calculate XOR results for every possible length up to the end of the collection, we are essentially storing results that would occupy space proportional to the number of possible subarrays. In the worst case, this would be on the order of N choose 2, which simplifies to N*(N+1)/2. Thus, the auxiliary space used is proportional to N squared, resulting in O(N^2) space complexity.

Optimal Solution

Approach

This problem has a hidden pattern related to how XOR works with numbers that follow a specific sequence. The best way to solve it is to figure out that pattern and use it to directly calculate the answer instead of doing the XOR operations step by step.

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

Notice that the numbers you're working with follow a simple sequence (starting with something, then increasing by two each time).
Figure out the mathematical pattern of XOR results when you do this sequence.
Depending on the starting number and the total count of numbers, the pattern lets you directly jump to the final XOR result without performing any individual XOR operations.
Use this pattern to calculate the final result as quickly as possible.

Code Implementation

def xor_operation_in_array(array_length, start_number):
    # If array_length is multiple of 4, result is start_number
    if array_length % 4 == 0:
        return start_number

    if array_length % 4 == 1:
        # If length % 4 == 1, result is start_number XOR 0
        return start_number

    if array_length % 4 == 2:
        # Result will be start_number XOR (start_number + 1 * 2)
        return start_number ^ (start_number + 1 * 2)

    # If length % 4 == 3, result is start_number XOR ... XOR (start_number + 3 * 2)
    return start_number ^ (start_number + 1 * 2) ^ (start_number + 2 * 2)

Big(O) Analysis

Time Complexity

O(1) – The solution leverages a mathematical pattern derived from the properties of XOR operations on a specific sequence. Instead of iterating or performing XOR operations, the algorithm directly calculates the result using a formula based on the input 'n' and the starting number. Therefore, the execution time is independent of the input size 'n', resulting in constant time complexity.

Space Complexity

O(1) – The described approach aims to find a mathematical pattern and directly calculate the final XOR result. It doesn't involve creating any auxiliary data structures such as arrays, lists, or hash maps to store intermediate values. The algorithm relies on a predetermined pattern based on the starting number and the count of numbers (N), but the identified pattern is calculated and returned, and does not require any extra memory that scales with N. Therefore, the space complexity is constant, independent of the input size N.

Edge Cases

Case	How to Handle
n is zero	Return 0 since the XOR of an empty set is defined as the identity element for XOR, which is 0.
n is one	Return 'start' directly as there is only one element in the array.
Large n causing integer overflow in calculation of nums[i]	Use a wider integer type (e.g., long) or check for overflow before the calculation of each element.
Large n leading to significant time complexity if not optimized	Optimize the XOR calculation using bit manipulation properties to reduce iterations instead of computing the full array.
Negative start value	The algorithm should handle negative start values correctly as it only involves addition and XOR operations.
start is close to Integer.MAX_VALUE/MIN_VALUE and n is large	Check for potential overflow when calculating 'start + 2 * i', consider using long to avoid it.
n is a very large number such that creating an array of size n is infeasible due to memory constraints	Avoid creating the array explicitly; instead calculate the XOR values iteratively without storing the array elements.
start is zero	The algorithm functions correctly with start equal to zero, calculating XOR of even numbers.

n is zero

How to Handle:

Return 0 since the XOR of an empty set is defined as the identity element for XOR, which is 0.

n is one

How to Handle:

Return 'start' directly as there is only one element in the array.

Large n causing integer overflow in calculation of nums[i]

How to Handle:

Use a wider integer type (e.g., long) or check for overflow before the calculation of each element.

Large n leading to significant time complexity if not optimized

How to Handle:

Optimize the XOR calculation using bit manipulation properties to reduce iterations instead of computing the full array.

Negative start value

How to Handle:

The algorithm should handle negative start values correctly as it only involves addition and XOR operations.

start is close to Integer.MAX_VALUE/MIN_VALUE and n is large

How to Handle:

Check for potential overflow when calculating 'start + 2 * i', consider using long to avoid it.

n is a very large number such that creating an array of size n is infeasible due to memory constraints

How to Handle:

Avoid creating the array explicitly; instead calculate the XOR values iteratively without storing the array elements.

start is zero

How to Handle:

The algorithm functions correctly with start equal to zero, calculating XOR of even numbers.

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 for `n` and `start`? Specifically, what are the minimum and maximum possible values for each?
Can `n` be zero? If so, what should the function return?
Is `start` guaranteed to be an integer, or could it be a floating-point number?
Could you provide a few example inputs and their corresponding expected outputs to confirm my understanding of the problem?
Are there any memory constraints I should be aware of given the potential size of `n`?

Brute Force Solution

Approach

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

Consider each element in the collection as a potential starting point.
From that starting point, calculate the XOR result by combining it with the next element, then with the next-next element, and so on until you reach the end of the collection.
Store the XOR result you calculated from each starting point and length.
Compare all the stored XOR results to see if any of them match the target value we're seeking.
If a match is found, that is our answer. If you exhaust all possibilities and no match is found, then no such result exists within the collection.

Code Implementation

def find_xor_brute_force(numbers, target):
    # Initialize the running total with zero as described in the instructions
    running_total = 0

    def explore_combinations(index, current_total):
        nonlocal running_total

        # Base case: If we reach the end of the array, check if target is found.
        if index == len(numbers):
            if current_total == target:
                return True
            else:
                return False

        # Explore including the current number in the XOR operation.
        if explore_combinations(index + 1, current_total ^ numbers[index]):
            return True

        # Explore excluding the current number in the XOR operation.
        # We need to explore every possible combination.
        if explore_combinations(index + 1, current_total):
            return True

        return False

    # Start exploring combinations from the beginning of the array.
    if explore_combinations(0, running_total):
        return True
    else:
        return False

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Notice that the numbers you're working with follow a simple sequence (starting with something, then increasing by two each time).
Figure out the mathematical pattern of XOR results when you do this sequence.
Depending on the starting number and the total count of numbers, the pattern lets you directly jump to the final XOR result without performing any individual XOR operations.
Use this pattern to calculate the final result as quickly as possible.

Code Implementation

def xor_operation_in_array(array_length, start_number):
    # If array_length is multiple of 4, result is start_number
    if array_length % 4 == 0:
        return start_number

    if array_length % 4 == 1:
        # If length % 4 == 1, result is start_number XOR 0
        return start_number

    if array_length % 4 == 2:
        # Result will be start_number XOR (start_number + 1 * 2)
        return start_number ^ (start_number + 1 * 2)

    # If length % 4 == 3, result is start_number XOR ... XOR (start_number + 3 * 2)
    return start_number ^ (start_number + 1 * 2) ^ (start_number + 2 * 2)

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
n is zero	Return 0 since the XOR of an empty set is defined as the identity element for XOR, which is 0.
n is one	Return 'start' directly as there is only one element in the array.
Large n causing integer overflow in calculation of nums[i]	Use a wider integer type (e.g., long) or check for overflow before the calculation of each element.
Large n leading to significant time complexity if not optimized	Optimize the XOR calculation using bit manipulation properties to reduce iterations instead of computing the full array.
Negative start value	The algorithm should handle negative start values correctly as it only involves addition and XOR operations.
start is close to Integer.MAX_VALUE/MIN_VALUE and n is large	Check for potential overflow when calculating 'start + 2 * i', consider using long to avoid it.
n is a very large number such that creating an array of size n is infeasible due to memory constraints	Avoid creating the array explicitly; instead calculate the XOR values iteratively without storing the array elements.
start is zero	The algorithm functions correctly with start equal to zero, calculating XOR of even numbers.

n is zero

How to Handle:

Return 0 since the XOR of an empty set is defined as the identity element for XOR, which is 0.

n is one

How to Handle:

Return 'start' directly as there is only one element in the array.

Large n causing integer overflow in calculation of nums[i]

How to Handle:

Use a wider integer type (e.g., long) or check for overflow before the calculation of each element.

Large n leading to significant time complexity if not optimized

How to Handle:

Optimize the XOR calculation using bit manipulation properties to reduce iterations instead of computing the full array.

Negative start value

How to Handle:

The algorithm should handle negative start values correctly as it only involves addition and XOR operations.

start is close to Integer.MAX_VALUE/MIN_VALUE and n is large

How to Handle:

Check for potential overflow when calculating 'start + 2 * i', consider using long to avoid it.

n is a very large number such that creating an array of size n is infeasible due to memory constraints

How to Handle:

Avoid creating the array explicitly; instead calculate the XOR values iteratively without storing the array elements.

start is zero

How to Handle:

The algorithm functions correctly with start equal to zero, calculating XOR of even numbers.

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