XOR Queries of a Subarray - Leetcode 1310

Introduction

In this article, we will explore the problem of processing XOR queries on a subarray, inspired by Leetcode Problem 1310. The goal is to efficiently calculate the XOR of segments of an array based on a set of queries.

Problem Description

Given an input array and an array of queries of the same size (though this doesn’t have to hold for all cases), we want to process each query, which specifies two indices, left and right. The task for each query is to calculate the XOR of all elements in the input array from index left to index right.

Understanding XOR

Before we dive into the solution, it's vital to understand how the XOR operation works:

Bitwise Operation: In binary representation, XOR outputs 1 for differing bits and 0 for identical bits. For example:
- 1 (0001) XOR 3 (0011) = 2 (0010)
Properties:
- Commutative and Associative: The order of operations in XOR doesn’t matter.
- Self-Inverse: Any number XORed with itself results in 0.

Given these properties, we can efficiently query for results.

Brute Force Approach

The naive or brute force approach involves directly iterating over the range specified by each query, resulting in a time complexity of O(Q * N), where Q is the number of queries and N is the size of the input array.

Optimized Approach with Prefix XOR

To optimize, we can use the concept of prefix XOR:

Definition: A prefix XOR array keeps track of the XOR of elements from the start of the array up to any index.
Computation:
- Initialize a prefix array with an additional leading zero.
- For each element in the input array, compute its prefix XOR by XORing the current element with the last computed value.
Querying:
- Given a query (left, right), we can easily compute the desired XOR using: [ \text(result) = \text(prefix)[right + 1] \oplus \text(prefix)[left] ]
- This allows each query to be answered in O(1) time after an O(N) time preprocessing step.

Implementation in Python

Here’s how we can implement both the prefix XOR computation and the querying in Python:

def xorQueries(arr, queries):
    n = len(arr)
    
    # Create a prefix XOR array
    prefix = [0] * (n + 1)
    for i in range(n):
        prefix[i + 1] = prefix[i] ^ arr[i]

    # Process each query
    result = []
    for left, right in queries:
        result.append(prefix[right + 1] ^ prefix[left])
    
    return result

In this implementation:

We create a prefix array of size n + 1 initialized with zeros.
We fill in the prefix array with cumulative XOR values.
Each query is answered in constant time using our previously calculated prefix XOR values.

Conclusion

Using the prefix XOR technique drastically reduces the time complexity of processing multiple XOR queries on an array. With this approach, the overall time complexity becomes O(N + Q), which is efficient for handling a high number of queries.

Keyword

XOR
Queries
Subarray
Prefix
Time Complexity
Algorithm

FAQ

Q1: What is XOR in the context of arrays?

A1: XOR (exclusive OR) is a bitwise operation that compares corresponding bits of two numbers and outputs 1 for differing bits and 0 for identical bits.

Q2: How do I optimize regular queries for subarrays?

A2: You can use the prefix XOR technique, which allows you to compute the XOR of any subarray in constant time after an initial linear time setup.

Q3: What is the time complexity of the optimized solution?

A3: The time complexity of the optimized solution is O(N + Q), where N is the size of the input array and Q is the number of queries.

Q4: Can I modify the input array while solving this problem?

A4: Yes, you can modify the input array for your calculations, but keep in mind it breaks the immutability principle often preferred in functional programming paradigms.

XOR Queries of a Subarray - Leetcode 1310 - Python