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Count Triplets That Can Form Two Arrays of Equal XOR

Medium

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Topics:

ArraysBit Manipulation

Given an array of integers arr.

We want to select three indices i, j and k where (0 <= i < j <= k < arr.length).

Let's define a and b as follows:

a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1]
b = arr[j] ^ arr[j + 1] ^ ... ^ arr[k]

Note that ^ denotes the bitwise-xor operation.

Return the number of triplets (i, j and k) Where a == b.

Example 1:

Input: arr = [2,3,1,6,7]
Output: 4
Explanation: The triplets are (0,1,2), (0,2,2), (2,3,4) and (2,4,4)

Example 2:

Input: arr = [1,1,1,1,1]
Output: 10

Constraints:

1 <= arr.length <= 300
1 <= arr[i] <= 10⁸

Try coding it in LeetCode

Count Triplets That Can Form Two Arrays of Equal XOR

Medium

Asked by:

16 views

Topics:

ArraysBit Manipulation

Given an array of integers arr.

We want to select three indices i, j and k where (0 <= i < j <= k < arr.length).

Let's define a and b as follows:

a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1]
b = arr[j] ^ arr[j + 1] ^ ... ^ arr[k]

Note that ^ denotes the bitwise-xor operation.

Return the number of triplets (i, j and k) Where a == b.

Example 1:

Input: arr = [2,3,1,6,7]
Output: 4
Explanation: The triplets are (0,1,2), (0,2,2), (2,3,4) and (2,4,4)

Example 2:

Input: arr = [1,1,1,1,1]
Output: 10

Constraints:

1 <= arr.length <= 300
1 <= arr[i] <= 10⁸

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 maximum size of the input array `arr`?
Can the elements in `arr` be zero or negative integers?
If no triplets satisfy the condition, what should I return?
Are there any constraints on the values within the array `arr` beyond being integers?
Is the input array guaranteed to be non-empty?

Brute Force Solution

Approach

The brute force method involves checking every single possible combination to find the answer. We try all possible splits in the given sequence, and check if that split produces two segments that satisfy our condition. If they do, we count it, and in the end we have our total.

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

Consider every possible starting point in the sequence.
From each starting point, consider every possible ending point to define the first segment.
Then, for each such segment, define the second segment that starts right after the first segment ends and goes until the end of the sequence.
Check if the two segments satisfy the given condition for equality.
If they do, increase the count of valid combinations.
Repeat this process by changing the start, end, and split points to check all possible combinations.
Finally, report the total number of combinations that satisfy the given equality condition.

Code Implementation

def count_triplets_brute_force(numbers):
    count_of_valid_triplets = 0
    array_length = len(numbers)

    for first_element in range(array_length):
        for second_element in range(first_element + 1, array_length):
            for third_element in range(second_element, array_length):
                first_array_xor_sum = 0

                # Calculate XOR sum of the first array
                for k in range(first_element, second_element):
                    first_array_xor_sum ^= numbers[k]

                second_array_xor_sum = 0

                # Calculate XOR sum of the second array
                for k in range(second_element, third_element + 1):
                    second_array_xor_sum ^= numbers[k]

                # Check if XOR sums are equal
                if first_array_xor_sum == second_array_xor_sum:
                    count_of_valid_triplets += 1

    return count_of_valid_triplets

Big(O) Analysis

Time Complexity

O(n^3) – The algorithm iterates through all possible starting points (i) in the array, which takes O(n) time. For each starting point, it iterates through all possible ending points (j) to define the first segment, also taking O(n) time. Then, for each such segment, it iterates through all possible ending points (k) of the second segment, which represents another O(n) time. Therefore, the overall time complexity is O(n * n * n), which simplifies to O(n^3).

Space Complexity

O(1) – The provided solution iterates through the array using nested loops and checks conditions based on XOR operations. No auxiliary data structures such as lists, hash maps, or recursion are used to store intermediate results. The algorithm only uses a few integer variables to keep track of loop indices and the count of valid triplets. Therefore, the auxiliary space used remains constant irrespective of the input size N, resulting in O(1) space complexity.

Optimal Solution

Approach

The key is recognizing the XOR property: if two subarrays have the same XOR value, their XOR is zero. We can use this to efficiently count valid triplets by focusing on finding when the XOR of a larger section of the data is zero.

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

Calculate the running XOR value as you move through the numbers.
Think of each location in your dataset as the potential end of a range.
At each location, check all previous locations to see if the XOR value between them is zero.
If the XOR value between two locations is zero, then any location between those two counts as a valid midpoint for forming two arrays with equal XOR.
Keep a running total of all the valid midpoints you find.

Code Implementation

def count_triplets(numbers) -> int:
    array_length = len(numbers)
    count = 0

    for i in range(array_length):
        xor_value = 0
        for j in range(i, array_length):
            xor_value ^= numbers[j]
            if xor_value == 0:
                # If the XOR is 0, all locations between i+1 and j are valid.
                count += j - i

    return count

Big(O) Analysis

Time Complexity

O(n²) – The dominant cost comes from the nested loop structure. The outer loop iterates through each of the n elements of the array. For each element in the outer loop, the inner loop iterates through all previous elements to check the XOR value. Thus, for an array of size n, the algorithm performs approximately n * (n - 1) / 2 XOR computations. This simplifies to a quadratic time complexity, O(n²).

Space Complexity

O(1) – The provided algorithm calculates the running XOR and iterates through the array using index variables. It doesn't create any auxiliary data structures that scale with the input size N (the number of elements in the input array). The algorithm only uses a few integer variables to store the running XOR and loop indices. Therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
Empty array	Return 0 immediately as no triplets can be formed.
Array with one element	Return 0 immediately as no triplets can be formed.
Array with two elements	Return 0 immediately as no triplets can be formed.
Array with three elements where all elements are the same (e.g., [1,1,1])	This case must be handled carefully as i, j, and k can be arranged in different ways to achieve XOR equality.
Large array size (close to the constraint limit)	The solution's time complexity should be carefully considered and optimized (O(n^2) or better is expected).
Array containing only zeros	This case will result in many valid triplets, requiring efficient counting.
Array with a long sequence of identical values	The algorithm should efficiently handle this without excessive iterations.
Integer overflow in XOR calculations (if applicable)	Ensure the data type used for XOR calculations can accommodate the maximum possible result.

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 maximum size of the input array `arr`?
Can the elements in `arr` be zero or negative integers?
If no triplets satisfy the condition, what should I return?
Are there any constraints on the values within the array `arr` beyond being integers?
Is the input array guaranteed to be non-empty?

Brute Force Solution

Approach

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

Consider every possible starting point in the sequence.
From each starting point, consider every possible ending point to define the first segment.
Then, for each such segment, define the second segment that starts right after the first segment ends and goes until the end of the sequence.
Check if the two segments satisfy the given condition for equality.
If they do, increase the count of valid combinations.
Repeat this process by changing the start, end, and split points to check all possible combinations.
Finally, report the total number of combinations that satisfy the given equality condition.

Code Implementation

def count_triplets_brute_force(numbers):
    count_of_valid_triplets = 0
    array_length = len(numbers)

    for first_element in range(array_length):
        for second_element in range(first_element + 1, array_length):
            for third_element in range(second_element, array_length):
                first_array_xor_sum = 0

                # Calculate XOR sum of the first array
                for k in range(first_element, second_element):
                    first_array_xor_sum ^= numbers[k]

                second_array_xor_sum = 0

                # Calculate XOR sum of the second array
                for k in range(second_element, third_element + 1):
                    second_array_xor_sum ^= numbers[k]

                # Check if XOR sums are equal
                if first_array_xor_sum == second_array_xor_sum:
                    count_of_valid_triplets += 1

    return count_of_valid_triplets

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Calculate the running XOR value as you move through the numbers.
Think of each location in your dataset as the potential end of a range.
At each location, check all previous locations to see if the XOR value between them is zero.
If the XOR value between two locations is zero, then any location between those two counts as a valid midpoint for forming two arrays with equal XOR.
Keep a running total of all the valid midpoints you find.

Code Implementation

def count_triplets(numbers) -> int:
    array_length = len(numbers)
    count = 0

    for i in range(array_length):
        xor_value = 0
        for j in range(i, array_length):
            xor_value ^= numbers[j]
            if xor_value == 0:
                # If the XOR is 0, all locations between i+1 and j are valid.
                count += j - i

    return count

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty array	Return 0 immediately as no triplets can be formed.
Array with one element	Return 0 immediately as no triplets can be formed.
Array with two elements	Return 0 immediately as no triplets can be formed.
Array with three elements where all elements are the same (e.g., [1,1,1])	This case must be handled carefully as i, j, and k can be arranged in different ways to achieve XOR equality.
Large array size (close to the constraint limit)	The solution's time complexity should be carefully considered and optimized (O(n^2) or better is expected).
Array containing only zeros	This case will result in many valid triplets, requiring efficient counting.
Array with a long sequence of identical values	The algorithm should efficiently handle this without excessive iterations.
Integer overflow in XOR calculations (if applicable)	Ensure the data type used for XOR calculations can accommodate the maximum possible result.