What is the time complexity of Count Triplets That Can Form Two Arrays of Equal XOR?

The optimal solution for Count Triplets That Can Form Two Arrays of Equal XOR runs in O(n) time and uses O(n) space. The time complexity is O(n) because we make a single pass through the array. The space complexity is O(n) due to the hashmap storing prefix XOR values.

What is the brute force solution for Count Triplets That Can Form Two Arrays of Equal XOR?

The brute force approach for Count Triplets That Can Form Two Arrays of Equal XOR is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking all possible triplet combinations of indices (i, j, k) to see if the XOR of the two subarrays is equal. This is straightforward but inefficient, as it checks every possible combination. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Triplets That Can Form Two Arrays of Equal XOR?

Count Triplets That Can Form Two Arrays of Equal XOR uses the following concepts: Array, Hash Table, Math, Bit Manipulation, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Triplets That Can Form Two Arrays of Equal XOR?

Always clarify the constraints and edge cases before diving into the solution. Think about how you can reduce the number of operations, especially with nested loops. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Count Triplets That Can Form Two Arrays of Equal XOR?

Not considering the boundaries of the indices correctly, leading to out-of-bounds errors. Failing to account for cases where the XOR of subarrays can be zero.

#1442

Count Triplets That Can Form Two Arrays of Equal XOR

Medium

ArrayHash TableMathBit ManipulationPrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution leverages the properties of XOR and prefix sums to reduce the time complexity. By using a hashmap to store the frequency of prefix XOR values, we can efficiently count valid triplets without needing nested loops.

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Algorithm

5 steps

1Step 1: Initialize a hashmap to store the frequency of prefix XOR values and a variable to keep track of the current prefix XOR.
2Step 2: Iterate through the array and calculate the prefix XOR while updating the hashmap.
3Step 3: For each prefix XOR value, check if it has been seen before. If it has, it means there are valid triplets that can be formed with the current index.
4Step 4: Count the number of valid triplets based on the frequency of the prefix XOR values.
5Step 5: Return the total count of valid triplets.

solution.py17 lines

1# Full working Python code
2
3def countTriplets(arr):
4    count = 0
5    prefix_xor = 0
6    freq = {0: 1}
7    for num in arr:
8        prefix_xor ^= num
9        if prefix_xor in freq:
10            count += freq[prefix_xor]
11            freq[prefix_xor] += 1
12        else:
13            freq[prefix_xor] = 1
14    return count
15
16# Example usage:
17print(countTriplets([2, 3, 1, 6, 7]))  # Output: 4

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Complexity note: The time complexity is O(n) because we make a single pass through the array. The space complexity is O(n) due to the hashmap storing prefix XOR values.

1Using prefix XOR helps in reducing the problem to counting occurrences of XOR values.
2The properties of XOR (a ^ a = 0) allow us to find subarrays with equal XOR efficiently.

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