How do you solve XOR Queries of a Subarray on LeetCode?

XOR Queries of a Subarray (LeetCode #1310) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + m) time complexity and O(n) space complexity. Key insight: Using prefix XOR allows for efficient range queries.

What is the time complexity of XOR Queries of a Subarray?

The optimal solution for XOR Queries of a Subarray runs in O(n + m) time and uses O(n) space. The time complexity is O(n) for building the prefix array and O(m) for processing m queries, making it efficient for large inputs.

What is the brute force solution for XOR Queries of a Subarray?

The brute force approach for XOR Queries of a Subarray is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach calculates the XOR for each query by iterating through the specified subarray. This method is straightforward but inefficient for large inputs. The optimal Optimal Solution approach improves this to O(n + m).

What algorithm or data structure is used to solve XOR Queries of a Subarray?

XOR Queries of a Subarray uses the following concepts: Array, Bit Manipulation, Prefix Sum. The recommended patterns to study are: Prefix Sum, Bit Manipulation.

What are the interview tips for XOR Queries of a Subarray?

Always consider edge cases, such as the smallest and largest possible indices. Explain your thought process clearly, especially when transitioning from brute force to optimal solutions. Practice writing code without an IDE to simulate real interview conditions.

What are common mistakes when solving XOR Queries of a Subarray?

Not handling the case where left index is 0 in the optimal solution. Forgetting that XOR is commutative and associative, which can simplify reasoning.

#1310

XOR Queries of a Subarray

Medium

ArrayBit ManipulationPrefix SumPrefix SumBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n + m)Space O(n)

By using a prefix XOR array, we can compute the XOR for any subarray in constant time after an initial linear time preprocessing step. This significantly reduces the time complexity.

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Algorithm

3 steps

1Step 1: Create a prefix XOR array where prefix[i] = arr[0] XOR arr[1] XOR ... XOR arr[i].
2Step 2: For each query, use the prefix XOR to compute the result as prefix[right] XOR prefix[left - 1] (handle left = 0 case separately).
3Step 3: Return the results list.

solution.py13 lines

1def xorQueries(arr, queries):
2    n = len(arr)
3    prefix = [0] * n
4    prefix[0] = arr[0]
5    for i in range(1, n):
6        prefix[i] = prefix[i - 1] ^ arr[i]
7    result = []
8    for left, right in queries:
9        if left == 0:
10            result.append(prefix[right])
11        else:
12            result.append(prefix[right] ^ prefix[left - 1])
13    return result

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Complexity note: The time complexity is O(n) for building the prefix array and O(m) for processing m queries, making it efficient for large inputs.

1Using prefix XOR allows for efficient range queries.
2XOR has properties that can simplify calculations, such as x ^ x = 0.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.